Limit of the infinite sequence involving $a_n, s_n$












3















Let $a_n=sqrt{n}$ and $s_n=a_1+a_2+ldots+a_n$. Find the limit : $$lim_{nto +infty}left[frac{frac{a_n}{s_n}}{-ln(1-frac{a_n}{s_n})}right]$$




I am hopelessly confused with this problem. Is L'Hopital or Cesaro-Stolz of any use here? Any hints. Thanks beforehand.






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    Let $a_n=sqrt{n}$ and $s_n=a_1+a_2+ldots+a_n$. Find the limit : $$lim_{nto +infty}left[frac{frac{a_n}{s_n}}{-ln(1-frac{a_n}{s_n})}right]$$




    I am hopelessly confused with this problem. Is L'Hopital or Cesaro-Stolz of any use here? Any hints. Thanks beforehand.






    real-analysis sequences-and-series







    share|cite|improve this question



























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      3








      3


      1






      Let $a_n=sqrt{n}$ and $s_n=a_1+a_2+ldots+a_n$. Find the limit : $$lim_{nto +infty}left[frac{frac{a_n}{s_n}}{-ln(1-frac{a_n}{s_n})}right]$$




      I am hopelessly confused with this problem. Is L'Hopital or Cesaro-Stolz of any use here? Any hints. Thanks beforehand.






      real-analysis sequences-and-series







      share|cite|improve this question
















      Let $a_n=sqrt{n}$ and $s_n=a_1+a_2+ldots+a_n$. Find the limit : $$lim_{nto +infty}left[frac{frac{a_n}{s_n}}{-ln(1-frac{a_n}{s_n})}right]$$




      I am hopelessly confused with this problem. Is L'Hopital or Cesaro-Stolz of any use here? Any hints. Thanks beforehand.






      real-analysis sequences-and-series




      real-analysis sequences-and-series






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      edited Nov 29 at 11:55









      Rebellos

      14.2k31244




      14.2k31244










      asked Nov 29 at 11:53









      vidyarthi

      2,8021831




      2,8021831






















          3 Answers
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          2














          We have that by Stolz-Cesàro:



          $$
          lim_{nto +infty} frac{a_n}{s_n}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{s_{n+1}-s_{n}}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{a_{n+1}}=lim_{nto +infty} frac{sqrt{n+1}-sqrt{n}}{sqrt{n+1}}=0
          $$



          Therefore:



          $$
          lim_{nto +infty}frac{frac{a_n}{s_n}}{-lnleft(1-frac{a_n}{s_n}right)}=lim_{xto 0}frac{x}{-lnleft(1-xright)}=1
          $$






          share|cite|improve this answer































            1














            Observe that $x_n=a_n/s_n$ converges to zero. Then, the limit becomes trivially solvable by Taylor expansion of the logarithm.



            Estimate:



            $$s_n=sum_{k=0}^n sqrt{n}approx int_{0}^n sqrt{n}dn=frac{2}{3}n^{3/2}$$



            (we are only interested in the power scaling - we need to see that $3/2>1/2$, so we don't need to talk about the error, but you can estimate the error from Euler-Maclaurin formula, which then proves that all the extra terms are negligible in the limit.



            So you have



            $$x_n=frac{a_n}{s_n}asympfrac{1}{n}underset{nto infty}{to} 0 $$



            By the way, you don't have to use integrals to show $a_n/s_nto 0$, I just find it elegant. You can use induction or upper bounds or anything similar.






            share|cite|improve this answer





















            • thanks, so the final limit is $1$, by the L'Hopital rule. Can we use L'Hopital here
              – vidyarthi
              Nov 29 at 12:02












            • Yes, after you reduce this to $lim_{xto0} frac{x}{-ln (1-x)}$, you can use whatever you want. Once you get some practice with these things, you remember $ln(1+x)approx x$ as one of the 5 most common approximations that you recognize in an instant, avoiding the need to calculate anything.
              – orion
              Nov 29 at 12:08



















            1














            Observe that $$lim_{tto 0} frac{t}{-ln (1-t)} =lim_{tto 0} frac{1}{(1-t)^{-1}} =lim_{tto 0} (1-t)=1$$
            So by Heine definition of limit we have that for any sequence $x_n $ tending to $0$ the following equality holds $$lim_{nto infty} frac{x_n}{-ln (1-x_n)}=1$$






            share|cite|improve this answer





















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              3 Answers
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              3 Answers
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              2














              We have that by Stolz-Cesàro:



              $$
              lim_{nto +infty} frac{a_n}{s_n}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{s_{n+1}-s_{n}}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{a_{n+1}}=lim_{nto +infty} frac{sqrt{n+1}-sqrt{n}}{sqrt{n+1}}=0
              $$



              Therefore:



              $$
              lim_{nto +infty}frac{frac{a_n}{s_n}}{-lnleft(1-frac{a_n}{s_n}right)}=lim_{xto 0}frac{x}{-lnleft(1-xright)}=1
              $$






              share|cite|improve this answer




























                2














                We have that by Stolz-Cesàro:



                $$
                lim_{nto +infty} frac{a_n}{s_n}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{s_{n+1}-s_{n}}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{a_{n+1}}=lim_{nto +infty} frac{sqrt{n+1}-sqrt{n}}{sqrt{n+1}}=0
                $$



                Therefore:



                $$
                lim_{nto +infty}frac{frac{a_n}{s_n}}{-lnleft(1-frac{a_n}{s_n}right)}=lim_{xto 0}frac{x}{-lnleft(1-xright)}=1
                $$






                share|cite|improve this answer


























                  2












                  2








                  2






                  We have that by Stolz-Cesàro:



                  $$
                  lim_{nto +infty} frac{a_n}{s_n}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{s_{n+1}-s_{n}}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{a_{n+1}}=lim_{nto +infty} frac{sqrt{n+1}-sqrt{n}}{sqrt{n+1}}=0
                  $$



                  Therefore:



                  $$
                  lim_{nto +infty}frac{frac{a_n}{s_n}}{-lnleft(1-frac{a_n}{s_n}right)}=lim_{xto 0}frac{x}{-lnleft(1-xright)}=1
                  $$






                  share|cite|improve this answer














                  We have that by Stolz-Cesàro:



                  $$
                  lim_{nto +infty} frac{a_n}{s_n}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{s_{n+1}-s_{n}}=lim_{nto +infty} frac{a_{n+1}-a_{n}}{a_{n+1}}=lim_{nto +infty} frac{sqrt{n+1}-sqrt{n}}{sqrt{n+1}}=0
                  $$



                  Therefore:



                  $$
                  lim_{nto +infty}frac{frac{a_n}{s_n}}{-lnleft(1-frac{a_n}{s_n}right)}=lim_{xto 0}frac{x}{-lnleft(1-xright)}=1
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 at 14:29









                  Mefitico

                  920117




                  920117










                  answered Nov 29 at 13:18









                  gimusi

                  1




                  1























                      1














                      Observe that $x_n=a_n/s_n$ converges to zero. Then, the limit becomes trivially solvable by Taylor expansion of the logarithm.



                      Estimate:



                      $$s_n=sum_{k=0}^n sqrt{n}approx int_{0}^n sqrt{n}dn=frac{2}{3}n^{3/2}$$



                      (we are only interested in the power scaling - we need to see that $3/2>1/2$, so we don't need to talk about the error, but you can estimate the error from Euler-Maclaurin formula, which then proves that all the extra terms are negligible in the limit.



                      So you have



                      $$x_n=frac{a_n}{s_n}asympfrac{1}{n}underset{nto infty}{to} 0 $$



                      By the way, you don't have to use integrals to show $a_n/s_nto 0$, I just find it elegant. You can use induction or upper bounds or anything similar.






                      share|cite|improve this answer





















                      • thanks, so the final limit is $1$, by the L'Hopital rule. Can we use L'Hopital here
                        – vidyarthi
                        Nov 29 at 12:02












                      • Yes, after you reduce this to $lim_{xto0} frac{x}{-ln (1-x)}$, you can use whatever you want. Once you get some practice with these things, you remember $ln(1+x)approx x$ as one of the 5 most common approximations that you recognize in an instant, avoiding the need to calculate anything.
                        – orion
                        Nov 29 at 12:08
















                      1














                      Observe that $x_n=a_n/s_n$ converges to zero. Then, the limit becomes trivially solvable by Taylor expansion of the logarithm.



                      Estimate:



                      $$s_n=sum_{k=0}^n sqrt{n}approx int_{0}^n sqrt{n}dn=frac{2}{3}n^{3/2}$$



                      (we are only interested in the power scaling - we need to see that $3/2>1/2$, so we don't need to talk about the error, but you can estimate the error from Euler-Maclaurin formula, which then proves that all the extra terms are negligible in the limit.



                      So you have



                      $$x_n=frac{a_n}{s_n}asympfrac{1}{n}underset{nto infty}{to} 0 $$



                      By the way, you don't have to use integrals to show $a_n/s_nto 0$, I just find it elegant. You can use induction or upper bounds or anything similar.






                      share|cite|improve this answer





















                      • thanks, so the final limit is $1$, by the L'Hopital rule. Can we use L'Hopital here
                        – vidyarthi
                        Nov 29 at 12:02












                      • Yes, after you reduce this to $lim_{xto0} frac{x}{-ln (1-x)}$, you can use whatever you want. Once you get some practice with these things, you remember $ln(1+x)approx x$ as one of the 5 most common approximations that you recognize in an instant, avoiding the need to calculate anything.
                        – orion
                        Nov 29 at 12:08














                      1












                      1








                      1






                      Observe that $x_n=a_n/s_n$ converges to zero. Then, the limit becomes trivially solvable by Taylor expansion of the logarithm.



                      Estimate:



                      $$s_n=sum_{k=0}^n sqrt{n}approx int_{0}^n sqrt{n}dn=frac{2}{3}n^{3/2}$$



                      (we are only interested in the power scaling - we need to see that $3/2>1/2$, so we don't need to talk about the error, but you can estimate the error from Euler-Maclaurin formula, which then proves that all the extra terms are negligible in the limit.



                      So you have



                      $$x_n=frac{a_n}{s_n}asympfrac{1}{n}underset{nto infty}{to} 0 $$



                      By the way, you don't have to use integrals to show $a_n/s_nto 0$, I just find it elegant. You can use induction or upper bounds or anything similar.






                      share|cite|improve this answer












                      Observe that $x_n=a_n/s_n$ converges to zero. Then, the limit becomes trivially solvable by Taylor expansion of the logarithm.



                      Estimate:



                      $$s_n=sum_{k=0}^n sqrt{n}approx int_{0}^n sqrt{n}dn=frac{2}{3}n^{3/2}$$



                      (we are only interested in the power scaling - we need to see that $3/2>1/2$, so we don't need to talk about the error, but you can estimate the error from Euler-Maclaurin formula, which then proves that all the extra terms are negligible in the limit.



                      So you have



                      $$x_n=frac{a_n}{s_n}asympfrac{1}{n}underset{nto infty}{to} 0 $$



                      By the way, you don't have to use integrals to show $a_n/s_nto 0$, I just find it elegant. You can use induction or upper bounds or anything similar.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 29 at 12:01









                      orion

                      12.9k11836




                      12.9k11836












                      • thanks, so the final limit is $1$, by the L'Hopital rule. Can we use L'Hopital here
                        – vidyarthi
                        Nov 29 at 12:02












                      • Yes, after you reduce this to $lim_{xto0} frac{x}{-ln (1-x)}$, you can use whatever you want. Once you get some practice with these things, you remember $ln(1+x)approx x$ as one of the 5 most common approximations that you recognize in an instant, avoiding the need to calculate anything.
                        – orion
                        Nov 29 at 12:08


















                      • thanks, so the final limit is $1$, by the L'Hopital rule. Can we use L'Hopital here
                        – vidyarthi
                        Nov 29 at 12:02












                      • Yes, after you reduce this to $lim_{xto0} frac{x}{-ln (1-x)}$, you can use whatever you want. Once you get some practice with these things, you remember $ln(1+x)approx x$ as one of the 5 most common approximations that you recognize in an instant, avoiding the need to calculate anything.
                        – orion
                        Nov 29 at 12:08
















                      thanks, so the final limit is $1$, by the L'Hopital rule. Can we use L'Hopital here
                      – vidyarthi
                      Nov 29 at 12:02






                      thanks, so the final limit is $1$, by the L'Hopital rule. Can we use L'Hopital here
                      – vidyarthi
                      Nov 29 at 12:02














                      Yes, after you reduce this to $lim_{xto0} frac{x}{-ln (1-x)}$, you can use whatever you want. Once you get some practice with these things, you remember $ln(1+x)approx x$ as one of the 5 most common approximations that you recognize in an instant, avoiding the need to calculate anything.
                      – orion
                      Nov 29 at 12:08




                      Yes, after you reduce this to $lim_{xto0} frac{x}{-ln (1-x)}$, you can use whatever you want. Once you get some practice with these things, you remember $ln(1+x)approx x$ as one of the 5 most common approximations that you recognize in an instant, avoiding the need to calculate anything.
                      – orion
                      Nov 29 at 12:08











                      1














                      Observe that $$lim_{tto 0} frac{t}{-ln (1-t)} =lim_{tto 0} frac{1}{(1-t)^{-1}} =lim_{tto 0} (1-t)=1$$
                      So by Heine definition of limit we have that for any sequence $x_n $ tending to $0$ the following equality holds $$lim_{nto infty} frac{x_n}{-ln (1-x_n)}=1$$






                      share|cite|improve this answer


























                        1














                        Observe that $$lim_{tto 0} frac{t}{-ln (1-t)} =lim_{tto 0} frac{1}{(1-t)^{-1}} =lim_{tto 0} (1-t)=1$$
                        So by Heine definition of limit we have that for any sequence $x_n $ tending to $0$ the following equality holds $$lim_{nto infty} frac{x_n}{-ln (1-x_n)}=1$$






                        share|cite|improve this answer
























                          1












                          1








                          1






                          Observe that $$lim_{tto 0} frac{t}{-ln (1-t)} =lim_{tto 0} frac{1}{(1-t)^{-1}} =lim_{tto 0} (1-t)=1$$
                          So by Heine definition of limit we have that for any sequence $x_n $ tending to $0$ the following equality holds $$lim_{nto infty} frac{x_n}{-ln (1-x_n)}=1$$






                          share|cite|improve this answer












                          Observe that $$lim_{tto 0} frac{t}{-ln (1-t)} =lim_{tto 0} frac{1}{(1-t)^{-1}} =lim_{tto 0} (1-t)=1$$
                          So by Heine definition of limit we have that for any sequence $x_n $ tending to $0$ the following equality holds $$lim_{nto infty} frac{x_n}{-ln (1-x_n)}=1$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 29 at 12:09









                          MotylaNogaTomkaMazura

                          6,577917




                          6,577917






























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