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











up vote
1
down vote

favorite
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




























    up vote
    1
    down vote

    favorite
    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


























      up vote
      1
      down vote

      favorite
      1









      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      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





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














           

          draft saved


          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004767%2fif-lambda-n-int-01-fracdt1tn-for-n-in-mathbbn-then%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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


















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004767%2fif-lambda-n-int-01-fracdt1tn-for-n-in-mathbbn-then%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d
          Read more

          SQL update select statement

          Image
          0 I'm trying to update an entry in my database (postgresql). I'm having a problem with updating the query: INSERT INTO prices (price, product_id) SELECT product_price, commn_id FROM products_temp as prod WHERE NOT EXISTS (SELECT * FROM prices od WHERE prod.commn_id = od.product_id) OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) Query works fine when i delete: OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) So it seems to me that it's looping through this operation. My question is, how can I fix it? sql postgresql
          Read more

          'app-layout' is not a known element: how to share Component with different Modules

          Image
          0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule
          Read more