square root n limit












3












$begingroup$


$$(x_{n})_{ngeq 2} x_{n}=sqrt[n]{1+sum_{k=2}^{n}(k-1)(k-1)!} $$



$$lim_{nrightarrow infty }frac{x_{n}}{n}=?$$



I'm trying to solve the sum I but I don't know how to rewrite it.Can you give me some hints?



edit:



The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $(n! - 1)$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion and make some simplifications I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?






calculus limits







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    $$(x_{n})_{ngeq 2} x_{n}=sqrt[n]{1+sum_{k=2}^{n}(k-1)(k-1)!} $$



    $$lim_{nrightarrow infty }frac{x_{n}}{n}=?$$



    I'm trying to solve the sum I but I don't know how to rewrite it.Can you give me some hints?



    edit:



    The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $(n! - 1)$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion and make some simplifications I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?






    calculus limits







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      $$(x_{n})_{ngeq 2} x_{n}=sqrt[n]{1+sum_{k=2}^{n}(k-1)(k-1)!} $$



      $$lim_{nrightarrow infty }frac{x_{n}}{n}=?$$



      I'm trying to solve the sum I but I don't know how to rewrite it.Can you give me some hints?



      edit:



      The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $(n! - 1)$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion and make some simplifications I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?






      calculus limits







      share|cite|improve this question











      $endgroup$




      $$(x_{n})_{ngeq 2} x_{n}=sqrt[n]{1+sum_{k=2}^{n}(k-1)(k-1)!} $$



      $$lim_{nrightarrow infty }frac{x_{n}}{n}=?$$



      I'm trying to solve the sum I but I don't know how to rewrite it.Can you give me some hints?



      edit:



      The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $(n! - 1)$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion and make some simplifications I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?






      calculus limits




      calculus limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago







      DaniVaja

















      asked 3 hours ago









      DaniVajaDaniVaja

      725




      725






















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          Very roughly, the sum will be dominated by the last term. You can then apply Stiriling's approximation to the last term
          $$(n-1)(n-1)!approx frac {(n-1)^n}{e^{n-1}}sqrt {2pi (n-1)}$$
          The $n^{th}$ root of this goes to $frac {n-1}e$ so the limit goes to $frac 1e$
          You need to justify the approximations made, particularly the one about considering only the last term of the sum.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
            $endgroup$
            – DaniVaja
            3 hours ago






          • 1




            $begingroup$
            As Vladimir said, you should ask the question you have. I would use induction. Show it is true for $k=2$, which is easy, then assume it is true up to $k$ and show it holds for $k+1$.
            $endgroup$
            – Ross Millikan
            1 hour ago



















          2












          $begingroup$

          Well, using Stirling's formula $m!=(m/e)^msqrt{2pi m}(1+o(1))$ as $mtoinfty$, we find that the radicand for large $k$ is $((n-1)/e)^{n-1} A_n$, where $1le A_nle Cn^{5/2}$ with some positive constant $C$ independent of $n$. Hence $lim_{ntoinfty}sqrt[n]{A_n}=1$, and we obtain
          $$
          lim_{ntoinfty} frac{x_n}n=lim_{ntoinfty}frac{(n-1)^{(n-1)/n}}{ne^{(n-1)/n}}=frac1e.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
            $endgroup$
            – DaniVaja
            2 hours ago






          • 1




            $begingroup$
            If that's your real problem, you should have written it in your question in the first place. We only try to answer questions posed and cannot generally read between the lines. Sorry, but I cannot be helpful further with your problem/
            $endgroup$
            – Vladimir
            2 hours ago



















          0












          $begingroup$

          Note that $(k-1)(k-1)!$ is very nearly $k(k-1)!$, which is $k!$. We'll use that:
          $$
          sum_{k=2}^n (k-1)(k-1)! + sum_{k=2}^n (k-1)! = sum_{k=2}^n (k-1+1)(k-1)! = sum_{k=2}^n k!
          $$
          Then (careful with the limits):
          $$
          sum_{k=2}^n (k-1)(k-1)! = sum_{k=2}^n k! - sum_{k=2}^n (k-1)! = sum_{k=2}^n k! - sum_{k=1}^{n-1} k! = n! - 1! tag{*}
          $$
          because almost all the terms cancel out.




          Edit: neater train of thought, but technically the same:

          Observe that $(k-1)(k-1)! = k! - (k-1)!$, and then follow with $(*)$.






          share|cite|improve this answer










          New contributor




          eudes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            Very roughly, the sum will be dominated by the last term. You can then apply Stiriling's approximation to the last term
            $$(n-1)(n-1)!approx frac {(n-1)^n}{e^{n-1}}sqrt {2pi (n-1)}$$
            The $n^{th}$ root of this goes to $frac {n-1}e$ so the limit goes to $frac 1e$
            You need to justify the approximations made, particularly the one about considering only the last term of the sum.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
              $endgroup$
              – DaniVaja
              3 hours ago






            • 1




              $begingroup$
              As Vladimir said, you should ask the question you have. I would use induction. Show it is true for $k=2$, which is easy, then assume it is true up to $k$ and show it holds for $k+1$.
              $endgroup$
              – Ross Millikan
              1 hour ago
















            5












            $begingroup$

            Very roughly, the sum will be dominated by the last term. You can then apply Stiriling's approximation to the last term
            $$(n-1)(n-1)!approx frac {(n-1)^n}{e^{n-1}}sqrt {2pi (n-1)}$$
            The $n^{th}$ root of this goes to $frac {n-1}e$ so the limit goes to $frac 1e$
            You need to justify the approximations made, particularly the one about considering only the last term of the sum.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
              $endgroup$
              – DaniVaja
              3 hours ago






            • 1




              $begingroup$
              As Vladimir said, you should ask the question you have. I would use induction. Show it is true for $k=2$, which is easy, then assume it is true up to $k$ and show it holds for $k+1$.
              $endgroup$
              – Ross Millikan
              1 hour ago














            5












            5








            5





            $begingroup$

            Very roughly, the sum will be dominated by the last term. You can then apply Stiriling's approximation to the last term
            $$(n-1)(n-1)!approx frac {(n-1)^n}{e^{n-1}}sqrt {2pi (n-1)}$$
            The $n^{th}$ root of this goes to $frac {n-1}e$ so the limit goes to $frac 1e$
            You need to justify the approximations made, particularly the one about considering only the last term of the sum.






            share|cite|improve this answer









            $endgroup$



            Very roughly, the sum will be dominated by the last term. You can then apply Stiriling's approximation to the last term
            $$(n-1)(n-1)!approx frac {(n-1)^n}{e^{n-1}}sqrt {2pi (n-1)}$$
            The $n^{th}$ root of this goes to $frac {n-1}e$ so the limit goes to $frac 1e$
            You need to justify the approximations made, particularly the one about considering only the last term of the sum.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Ross MillikanRoss Millikan

            295k23198371




            295k23198371












            • $begingroup$
              Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
              $endgroup$
              – DaniVaja
              3 hours ago






            • 1




              $begingroup$
              As Vladimir said, you should ask the question you have. I would use induction. Show it is true for $k=2$, which is easy, then assume it is true up to $k$ and show it holds for $k+1$.
              $endgroup$
              – Ross Millikan
              1 hour ago


















            • $begingroup$
              Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
              $endgroup$
              – DaniVaja
              3 hours ago






            • 1




              $begingroup$
              As Vladimir said, you should ask the question you have. I would use induction. Show it is true for $k=2$, which is easy, then assume it is true up to $k$ and show it holds for $k+1$.
              $endgroup$
              – Ross Millikan
              1 hour ago
















            $begingroup$
            Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
            $endgroup$
            – DaniVaja
            3 hours ago




            $begingroup$
            Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
            $endgroup$
            – DaniVaja
            3 hours ago




            1




            1




            $begingroup$
            As Vladimir said, you should ask the question you have. I would use induction. Show it is true for $k=2$, which is easy, then assume it is true up to $k$ and show it holds for $k+1$.
            $endgroup$
            – Ross Millikan
            1 hour ago




            $begingroup$
            As Vladimir said, you should ask the question you have. I would use induction. Show it is true for $k=2$, which is easy, then assume it is true up to $k$ and show it holds for $k+1$.
            $endgroup$
            – Ross Millikan
            1 hour ago











            2












            $begingroup$

            Well, using Stirling's formula $m!=(m/e)^msqrt{2pi m}(1+o(1))$ as $mtoinfty$, we find that the radicand for large $k$ is $((n-1)/e)^{n-1} A_n$, where $1le A_nle Cn^{5/2}$ with some positive constant $C$ independent of $n$. Hence $lim_{ntoinfty}sqrt[n]{A_n}=1$, and we obtain
            $$
            lim_{ntoinfty} frac{x_n}n=lim_{ntoinfty}frac{(n-1)^{(n-1)/n}}{ne^{(n-1)/n}}=frac1e.
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
              $endgroup$
              – DaniVaja
              2 hours ago






            • 1




              $begingroup$
              If that's your real problem, you should have written it in your question in the first place. We only try to answer questions posed and cannot generally read between the lines. Sorry, but I cannot be helpful further with your problem/
              $endgroup$
              – Vladimir
              2 hours ago
















            2












            $begingroup$

            Well, using Stirling's formula $m!=(m/e)^msqrt{2pi m}(1+o(1))$ as $mtoinfty$, we find that the radicand for large $k$ is $((n-1)/e)^{n-1} A_n$, where $1le A_nle Cn^{5/2}$ with some positive constant $C$ independent of $n$. Hence $lim_{ntoinfty}sqrt[n]{A_n}=1$, and we obtain
            $$
            lim_{ntoinfty} frac{x_n}n=lim_{ntoinfty}frac{(n-1)^{(n-1)/n}}{ne^{(n-1)/n}}=frac1e.
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
              $endgroup$
              – DaniVaja
              2 hours ago






            • 1




              $begingroup$
              If that's your real problem, you should have written it in your question in the first place. We only try to answer questions posed and cannot generally read between the lines. Sorry, but I cannot be helpful further with your problem/
              $endgroup$
              – Vladimir
              2 hours ago














            2












            2








            2





            $begingroup$

            Well, using Stirling's formula $m!=(m/e)^msqrt{2pi m}(1+o(1))$ as $mtoinfty$, we find that the radicand for large $k$ is $((n-1)/e)^{n-1} A_n$, where $1le A_nle Cn^{5/2}$ with some positive constant $C$ independent of $n$. Hence $lim_{ntoinfty}sqrt[n]{A_n}=1$, and we obtain
            $$
            lim_{ntoinfty} frac{x_n}n=lim_{ntoinfty}frac{(n-1)^{(n-1)/n}}{ne^{(n-1)/n}}=frac1e.
            $$






            share|cite|improve this answer









            $endgroup$



            Well, using Stirling's formula $m!=(m/e)^msqrt{2pi m}(1+o(1))$ as $mtoinfty$, we find that the radicand for large $k$ is $((n-1)/e)^{n-1} A_n$, where $1le A_nle Cn^{5/2}$ with some positive constant $C$ independent of $n$. Hence $lim_{ntoinfty}sqrt[n]{A_n}=1$, and we obtain
            $$
            lim_{ntoinfty} frac{x_n}n=lim_{ntoinfty}frac{(n-1)^{(n-1)/n}}{ne^{(n-1)/n}}=frac1e.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            VladimirVladimir

            5,322618




            5,322618












            • $begingroup$
              Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
              $endgroup$
              – DaniVaja
              2 hours ago






            • 1




              $begingroup$
              If that's your real problem, you should have written it in your question in the first place. We only try to answer questions posed and cannot generally read between the lines. Sorry, but I cannot be helpful further with your problem/
              $endgroup$
              – Vladimir
              2 hours ago


















            • $begingroup$
              Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
              $endgroup$
              – DaniVaja
              2 hours ago






            • 1




              $begingroup$
              If that's your real problem, you should have written it in your question in the first place. We only try to answer questions posed and cannot generally read between the lines. Sorry, but I cannot be helpful further with your problem/
              $endgroup$
              – Vladimir
              2 hours ago
















            $begingroup$
            Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
            $endgroup$
            – DaniVaja
            2 hours ago




            $begingroup$
            Hi, thank you the response.The sum $sum_{k=2}^{n}(k-1)(k-1)!$ should be $n! - 1$ so $x_{n}=sqrt[n]{n!}$.The limit would be $lim_{nrightarrow infty } sqrt[n]{frac{n!}{n^{n}}}$ .If I use cauchy d'alembert criterion I would get $e^{-1}$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?
            $endgroup$
            – DaniVaja
            2 hours ago




            1




            1




            $begingroup$
            If that's your real problem, you should have written it in your question in the first place. We only try to answer questions posed and cannot generally read between the lines. Sorry, but I cannot be helpful further with your problem/
            $endgroup$
            – Vladimir
            2 hours ago




            $begingroup$
            If that's your real problem, you should have written it in your question in the first place. We only try to answer questions posed and cannot generally read between the lines. Sorry, but I cannot be helpful further with your problem/
            $endgroup$
            – Vladimir
            2 hours ago











            0












            $begingroup$

            Note that $(k-1)(k-1)!$ is very nearly $k(k-1)!$, which is $k!$. We'll use that:
            $$
            sum_{k=2}^n (k-1)(k-1)! + sum_{k=2}^n (k-1)! = sum_{k=2}^n (k-1+1)(k-1)! = sum_{k=2}^n k!
            $$
            Then (careful with the limits):
            $$
            sum_{k=2}^n (k-1)(k-1)! = sum_{k=2}^n k! - sum_{k=2}^n (k-1)! = sum_{k=2}^n k! - sum_{k=1}^{n-1} k! = n! - 1! tag{*}
            $$
            because almost all the terms cancel out.




            Edit: neater train of thought, but technically the same:

            Observe that $(k-1)(k-1)! = k! - (k-1)!$, and then follow with $(*)$.






            share|cite|improve this answer










            New contributor




            eudes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$


















              0












              $begingroup$

              Note that $(k-1)(k-1)!$ is very nearly $k(k-1)!$, which is $k!$. We'll use that:
              $$
              sum_{k=2}^n (k-1)(k-1)! + sum_{k=2}^n (k-1)! = sum_{k=2}^n (k-1+1)(k-1)! = sum_{k=2}^n k!
              $$
              Then (careful with the limits):
              $$
              sum_{k=2}^n (k-1)(k-1)! = sum_{k=2}^n k! - sum_{k=2}^n (k-1)! = sum_{k=2}^n k! - sum_{k=1}^{n-1} k! = n! - 1! tag{*}
              $$
              because almost all the terms cancel out.




              Edit: neater train of thought, but technically the same:

              Observe that $(k-1)(k-1)! = k! - (k-1)!$, and then follow with $(*)$.






              share|cite|improve this answer










              New contributor




              eudes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$
















                0












                0








                0





                $begingroup$

                Note that $(k-1)(k-1)!$ is very nearly $k(k-1)!$, which is $k!$. We'll use that:
                $$
                sum_{k=2}^n (k-1)(k-1)! + sum_{k=2}^n (k-1)! = sum_{k=2}^n (k-1+1)(k-1)! = sum_{k=2}^n k!
                $$
                Then (careful with the limits):
                $$
                sum_{k=2}^n (k-1)(k-1)! = sum_{k=2}^n k! - sum_{k=2}^n (k-1)! = sum_{k=2}^n k! - sum_{k=1}^{n-1} k! = n! - 1! tag{*}
                $$
                because almost all the terms cancel out.




                Edit: neater train of thought, but technically the same:

                Observe that $(k-1)(k-1)! = k! - (k-1)!$, and then follow with $(*)$.






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                $endgroup$



                Note that $(k-1)(k-1)!$ is very nearly $k(k-1)!$, which is $k!$. We'll use that:
                $$
                sum_{k=2}^n (k-1)(k-1)! + sum_{k=2}^n (k-1)! = sum_{k=2}^n (k-1+1)(k-1)! = sum_{k=2}^n k!
                $$
                Then (careful with the limits):
                $$
                sum_{k=2}^n (k-1)(k-1)! = sum_{k=2}^n k! - sum_{k=2}^n (k-1)! = sum_{k=2}^n k! - sum_{k=1}^{n-1} k! = n! - 1! tag{*}
                $$
                because almost all the terms cancel out.




                Edit: neater train of thought, but technically the same:

                Observe that $(k-1)(k-1)! = k! - (k-1)!$, and then follow with $(*)$.







                share|cite|improve this answer










                New contributor




                eudes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago





















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                answered 1 hour ago









                eudeseudes

                11




                11




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