If $lambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$, for $n in mathbb{N}$, then $,lim_{n to infty}...











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If $displaystylelambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$ for $n in mathbb{N}$. Then prove that $lim_{n to infty} (lambda_{n})^{1/n}=1.$



$$lambda_n=int_{0}^{1} frac{dt}{(1+t)^n}= frac{2^{1-n}}{1-n}-frac{1}{1-n}$$



Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.






calculus real-analysis limits definite-integrals limits-without-lhopital







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    If $displaystylelambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$ for $n in mathbb{N}$. Then prove that $lim_{n to infty} (lambda_{n})^{1/n}=1.$



    $$lambda_n=int_{0}^{1} frac{dt}{(1+t)^n}= frac{2^{1-n}}{1-n}-frac{1}{1-n}$$



    Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.






    calculus real-analysis limits definite-integrals limits-without-lhopital







    share|cite|improve this question


























      up vote
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      up vote
      1
      down vote

      favorite
      1






      1





      If $displaystylelambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$ for $n in mathbb{N}$. Then prove that $lim_{n to infty} (lambda_{n})^{1/n}=1.$



      $$lambda_n=int_{0}^{1} frac{dt}{(1+t)^n}= frac{2^{1-n}}{1-n}-frac{1}{1-n}$$



      Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.






      calculus real-analysis limits definite-integrals limits-without-lhopital







      share|cite|improve this question















      If $displaystylelambda_n = int_{0}^{1} frac{dt}{(1+t)^n}$ for $n in mathbb{N}$. Then prove that $lim_{n to infty} (lambda_{n})^{1/n}=1.$



      $$lambda_n=int_{0}^{1} frac{dt}{(1+t)^n}= frac{2^{1-n}}{1-n}-frac{1}{1-n}$$



      Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.






      calculus real-analysis limits definite-integrals limits-without-lhopital




      calculus real-analysis limits definite-integrals limits-without-lhopital






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      edited 17 hours ago









      Yiorgos S. Smyrlis

      61.4k1383161




      61.4k1383161










      asked 18 hours ago









      ramanujan

      640713




      640713






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          Actually,
          $$
          int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
          $$
          and hence, for all $n>1$
          $$
          frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
          $$
          Next, observe that
          $$
          lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
          $$






          share|cite|improve this answer





















          • thank you. I wonder how such tricks come in your mind.
            – ramanujan
            18 hours ago






          • 1




            @ramanujan: the old sage said Calculus is all about elementary inequalities.
            – Jack D'Aurizio
            15 hours ago


















          up vote
          2
          down vote













          In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,



          $$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
          by the inequality between means.






          share|cite|improve this answer





















          • Thanks for general result
            – ramanujan
            15 hours ago











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          Actually,
          $$
          int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
          $$
          and hence, for all $n>1$
          $$
          frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
          $$
          Next, observe that
          $$
          lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
          $$






          share|cite|improve this answer





















          • thank you. I wonder how such tricks come in your mind.
            – ramanujan
            18 hours ago






          • 1




            @ramanujan: the old sage said Calculus is all about elementary inequalities.
            – Jack D'Aurizio
            15 hours ago















          up vote
          4
          down vote



          accepted










          Actually,
          $$
          int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
          $$
          and hence, for all $n>1$
          $$
          frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
          $$
          Next, observe that
          $$
          lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
          $$






          share|cite|improve this answer





















          • thank you. I wonder how such tricks come in your mind.
            – ramanujan
            18 hours ago






          • 1




            @ramanujan: the old sage said Calculus is all about elementary inequalities.
            – Jack D'Aurizio
            15 hours ago













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Actually,
          $$
          int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
          $$
          and hence, for all $n>1$
          $$
          frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
          $$
          Next, observe that
          $$
          lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
          $$






          share|cite|improve this answer












          Actually,
          $$
          int_0^1 frac{dt}{(1+t)^n}=left.frac{1}{1-n}frac{1}{(1+t)^{n-1}},right|_0^1=frac{1}{n-1}-frac{2^{-n+1}}{n-1}
          $$
          and hence, for all $n>1$
          $$
          frac{1}{2(n-1)}<int_0^1 frac{dt}{(1+t)^n}<frac{1}{n-1}.
          $$
          Next, observe that
          $$
          lim_{ntoinfty}left(frac{1}{2(n-1)}right)^{1/n}=lim_{ntoinfty}left(frac{1}{n-1}right)^{1/n}=1.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 18 hours ago









          Yiorgos S. Smyrlis

          61.4k1383161




          61.4k1383161












          • thank you. I wonder how such tricks come in your mind.
            – ramanujan
            18 hours ago






          • 1




            @ramanujan: the old sage said Calculus is all about elementary inequalities.
            – Jack D'Aurizio
            15 hours ago


















          • thank you. I wonder how such tricks come in your mind.
            – ramanujan
            18 hours ago






          • 1




            @ramanujan: the old sage said Calculus is all about elementary inequalities.
            – Jack D'Aurizio
            15 hours ago
















          thank you. I wonder how such tricks come in your mind.
          – ramanujan
          18 hours ago




          thank you. I wonder how such tricks come in your mind.
          – ramanujan
          18 hours ago




          1




          1




          @ramanujan: the old sage said Calculus is all about elementary inequalities.
          – Jack D'Aurizio
          15 hours ago




          @ramanujan: the old sage said Calculus is all about elementary inequalities.
          – Jack D'Aurizio
          15 hours ago










          up vote
          2
          down vote













          In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,



          $$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
          by the inequality between means.






          share|cite|improve this answer





















          • Thanks for general result
            – ramanujan
            15 hours ago















          up vote
          2
          down vote













          In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,



          $$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
          by the inequality between means.






          share|cite|improve this answer





















          • Thanks for general result
            – ramanujan
            15 hours ago













          up vote
          2
          down vote










          up vote
          2
          down vote









          In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,



          $$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
          by the inequality between means.






          share|cite|improve this answer












          In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$,



          $$ lim_{nto +infty}sqrt[n]{int_{0}^{1}f(x)^n,dx} = max_{xin[0,1]}f(x) $$
          by the inequality between means.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 15 hours ago









          Jack D'Aurizio

          282k33274653




          282k33274653












          • Thanks for general result
            – ramanujan
            15 hours ago


















          • Thanks for general result
            – ramanujan
            15 hours ago
















          Thanks for general result
          – ramanujan
          15 hours ago




          Thanks for general result
          – ramanujan
          15 hours ago


















           

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