Real Analysis, $limlimits_{nrightarrowinfty} int_{0}^{1} frac{e^{-nt}-(1-t)^n}{t} dt$












4














I was trying to compute $limlimits_{nrightarrowinfty} int_{0}^{1} frac{e^{-nt}-(1-t)^n}{t} dt$ using Lebesgue's dominated convergence THM, but I can't exactly figure out how to do. I mean, I managed to prove that each the integrand function $f_n(t)$ is less or equal than $g_n(t)=e^{-nt}sqrt{n}e^frac{1}{sqrt{n}}, forall ninmathbb{N}$. And since we are dealing with positive functions and $int_{0}^{1} g_n(t)dtleqfrac{e}{sqrt{n}}overset{mathrm{nrightarrowinfty}}{rightarrow}0$, I can deduce that the original limit is $0$.




Now, I was just wondering if anyone is able to show analytically that there exists a function in $mathcal{L}^1$ which dominates all the $f_n$ in order to apply Lebesgue's dominated convergence THM. I made some attempts, but I failed.




Thanks in advance,
a humble half-mathematician.






real-analysis integration limits convergence definite-integrals







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    I was trying to compute $limlimits_{nrightarrowinfty} int_{0}^{1} frac{e^{-nt}-(1-t)^n}{t} dt$ using Lebesgue's dominated convergence THM, but I can't exactly figure out how to do. I mean, I managed to prove that each the integrand function $f_n(t)$ is less or equal than $g_n(t)=e^{-nt}sqrt{n}e^frac{1}{sqrt{n}}, forall ninmathbb{N}$. And since we are dealing with positive functions and $int_{0}^{1} g_n(t)dtleqfrac{e}{sqrt{n}}overset{mathrm{nrightarrowinfty}}{rightarrow}0$, I can deduce that the original limit is $0$.




    Now, I was just wondering if anyone is able to show analytically that there exists a function in $mathcal{L}^1$ which dominates all the $f_n$ in order to apply Lebesgue's dominated convergence THM. I made some attempts, but I failed.




    Thanks in advance,
    a humble half-mathematician.






    real-analysis integration limits convergence definite-integrals







    share|cite|improve this question



























      4












      4








      4


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      I was trying to compute $limlimits_{nrightarrowinfty} int_{0}^{1} frac{e^{-nt}-(1-t)^n}{t} dt$ using Lebesgue's dominated convergence THM, but I can't exactly figure out how to do. I mean, I managed to prove that each the integrand function $f_n(t)$ is less or equal than $g_n(t)=e^{-nt}sqrt{n}e^frac{1}{sqrt{n}}, forall ninmathbb{N}$. And since we are dealing with positive functions and $int_{0}^{1} g_n(t)dtleqfrac{e}{sqrt{n}}overset{mathrm{nrightarrowinfty}}{rightarrow}0$, I can deduce that the original limit is $0$.




      Now, I was just wondering if anyone is able to show analytically that there exists a function in $mathcal{L}^1$ which dominates all the $f_n$ in order to apply Lebesgue's dominated convergence THM. I made some attempts, but I failed.




      Thanks in advance,
      a humble half-mathematician.






      real-analysis integration limits convergence definite-integrals







      share|cite|improve this question















      I was trying to compute $limlimits_{nrightarrowinfty} int_{0}^{1} frac{e^{-nt}-(1-t)^n}{t} dt$ using Lebesgue's dominated convergence THM, but I can't exactly figure out how to do. I mean, I managed to prove that each the integrand function $f_n(t)$ is less or equal than $g_n(t)=e^{-nt}sqrt{n}e^frac{1}{sqrt{n}}, forall ninmathbb{N}$. And since we are dealing with positive functions and $int_{0}^{1} g_n(t)dtleqfrac{e}{sqrt{n}}overset{mathrm{nrightarrowinfty}}{rightarrow}0$, I can deduce that the original limit is $0$.




      Now, I was just wondering if anyone is able to show analytically that there exists a function in $mathcal{L}^1$ which dominates all the $f_n$ in order to apply Lebesgue's dominated convergence THM. I made some attempts, but I failed.




      Thanks in advance,
      a humble half-mathematician.






      real-analysis integration limits convergence definite-integrals




      real-analysis integration limits convergence definite-integrals






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      share|cite|improve this question













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      edited Dec 8 at 11:46









      Batominovski

      33.7k33292




      33.7k33292










      asked Dec 8 at 10:24









      piotor

      385




      385






















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          We have
          $$exp(-nt)-(1-t)^n=exp(-nt),Big(1-big((1-t)exp(t)big)^nBig),.tag{*}$$
          By Bernoulli's Inequaliy,
          $$big((1-t)exp(t)big)^ngeq 1+n,big((1-t)exp(t)-1big),.$$
          Therefore,
          $$exp(-nt)-(1-t)^nleq n,exp(-nt),big(1-(1-t)exp(t)big),.$$
          By taking derivative with respect to $n$, we can show that
          $$n,exp(-nt)leq frac{1}{text{e},t}text{ for all }t>0text{ and positive integers }n,.$$
          Furthermore, we have
          $$(1-t),exp(t)=1-sum_{k=1}^infty,left(frac{k-1}{k!}right),t^kgeq 1-t^2,sum_{k=2}^infty,frac{k-1}{k!}=1-t^2$$
          for all $tin[0,1]$.



          By (*), we conclude that
          $$f_n(t):=frac{exp(-nt)-(1-t)^n}{t}leq frac{1}{text{e},t^2},big(1-(1-t^2)big)=frac{1}{text{e}}$$
          for every $tin (0,1]$ and every positive integer $n$ (where the only equality case is when $n=1$ and $t=1$). Therefore, the constant function $fequiv dfrac{1}{text{e}}$ dominates $f_n$ for all $n=1,2,3,ldots$.






          share|cite|improve this answer





















          • Absolutely brilliant. Thanks a lot.
            – piotor
            Dec 8 at 18:45











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          3














          We have
          $$exp(-nt)-(1-t)^n=exp(-nt),Big(1-big((1-t)exp(t)big)^nBig),.tag{*}$$
          By Bernoulli's Inequaliy,
          $$big((1-t)exp(t)big)^ngeq 1+n,big((1-t)exp(t)-1big),.$$
          Therefore,
          $$exp(-nt)-(1-t)^nleq n,exp(-nt),big(1-(1-t)exp(t)big),.$$
          By taking derivative with respect to $n$, we can show that
          $$n,exp(-nt)leq frac{1}{text{e},t}text{ for all }t>0text{ and positive integers }n,.$$
          Furthermore, we have
          $$(1-t),exp(t)=1-sum_{k=1}^infty,left(frac{k-1}{k!}right),t^kgeq 1-t^2,sum_{k=2}^infty,frac{k-1}{k!}=1-t^2$$
          for all $tin[0,1]$.



          By (*), we conclude that
          $$f_n(t):=frac{exp(-nt)-(1-t)^n}{t}leq frac{1}{text{e},t^2},big(1-(1-t^2)big)=frac{1}{text{e}}$$
          for every $tin (0,1]$ and every positive integer $n$ (where the only equality case is when $n=1$ and $t=1$). Therefore, the constant function $fequiv dfrac{1}{text{e}}$ dominates $f_n$ for all $n=1,2,3,ldots$.






          share|cite|improve this answer





















          • Absolutely brilliant. Thanks a lot.
            – piotor
            Dec 8 at 18:45
















          3














          We have
          $$exp(-nt)-(1-t)^n=exp(-nt),Big(1-big((1-t)exp(t)big)^nBig),.tag{*}$$
          By Bernoulli's Inequaliy,
          $$big((1-t)exp(t)big)^ngeq 1+n,big((1-t)exp(t)-1big),.$$
          Therefore,
          $$exp(-nt)-(1-t)^nleq n,exp(-nt),big(1-(1-t)exp(t)big),.$$
          By taking derivative with respect to $n$, we can show that
          $$n,exp(-nt)leq frac{1}{text{e},t}text{ for all }t>0text{ and positive integers }n,.$$
          Furthermore, we have
          $$(1-t),exp(t)=1-sum_{k=1}^infty,left(frac{k-1}{k!}right),t^kgeq 1-t^2,sum_{k=2}^infty,frac{k-1}{k!}=1-t^2$$
          for all $tin[0,1]$.



          By (*), we conclude that
          $$f_n(t):=frac{exp(-nt)-(1-t)^n}{t}leq frac{1}{text{e},t^2},big(1-(1-t^2)big)=frac{1}{text{e}}$$
          for every $tin (0,1]$ and every positive integer $n$ (where the only equality case is when $n=1$ and $t=1$). Therefore, the constant function $fequiv dfrac{1}{text{e}}$ dominates $f_n$ for all $n=1,2,3,ldots$.






          share|cite|improve this answer





















          • Absolutely brilliant. Thanks a lot.
            – piotor
            Dec 8 at 18:45














          3












          3








          3






          We have
          $$exp(-nt)-(1-t)^n=exp(-nt),Big(1-big((1-t)exp(t)big)^nBig),.tag{*}$$
          By Bernoulli's Inequaliy,
          $$big((1-t)exp(t)big)^ngeq 1+n,big((1-t)exp(t)-1big),.$$
          Therefore,
          $$exp(-nt)-(1-t)^nleq n,exp(-nt),big(1-(1-t)exp(t)big),.$$
          By taking derivative with respect to $n$, we can show that
          $$n,exp(-nt)leq frac{1}{text{e},t}text{ for all }t>0text{ and positive integers }n,.$$
          Furthermore, we have
          $$(1-t),exp(t)=1-sum_{k=1}^infty,left(frac{k-1}{k!}right),t^kgeq 1-t^2,sum_{k=2}^infty,frac{k-1}{k!}=1-t^2$$
          for all $tin[0,1]$.



          By (*), we conclude that
          $$f_n(t):=frac{exp(-nt)-(1-t)^n}{t}leq frac{1}{text{e},t^2},big(1-(1-t^2)big)=frac{1}{text{e}}$$
          for every $tin (0,1]$ and every positive integer $n$ (where the only equality case is when $n=1$ and $t=1$). Therefore, the constant function $fequiv dfrac{1}{text{e}}$ dominates $f_n$ for all $n=1,2,3,ldots$.






          share|cite|improve this answer












          We have
          $$exp(-nt)-(1-t)^n=exp(-nt),Big(1-big((1-t)exp(t)big)^nBig),.tag{*}$$
          By Bernoulli's Inequaliy,
          $$big((1-t)exp(t)big)^ngeq 1+n,big((1-t)exp(t)-1big),.$$
          Therefore,
          $$exp(-nt)-(1-t)^nleq n,exp(-nt),big(1-(1-t)exp(t)big),.$$
          By taking derivative with respect to $n$, we can show that
          $$n,exp(-nt)leq frac{1}{text{e},t}text{ for all }t>0text{ and positive integers }n,.$$
          Furthermore, we have
          $$(1-t),exp(t)=1-sum_{k=1}^infty,left(frac{k-1}{k!}right),t^kgeq 1-t^2,sum_{k=2}^infty,frac{k-1}{k!}=1-t^2$$
          for all $tin[0,1]$.



          By (*), we conclude that
          $$f_n(t):=frac{exp(-nt)-(1-t)^n}{t}leq frac{1}{text{e},t^2},big(1-(1-t^2)big)=frac{1}{text{e}}$$
          for every $tin (0,1]$ and every positive integer $n$ (where the only equality case is when $n=1$ and $t=1$). Therefore, the constant function $fequiv dfrac{1}{text{e}}$ dominates $f_n$ for all $n=1,2,3,ldots$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 at 11:39









          Batominovski

          33.7k33292




          33.7k33292












          • Absolutely brilliant. Thanks a lot.
            – piotor
            Dec 8 at 18:45


















          • Absolutely brilliant. Thanks a lot.
            – piotor
            Dec 8 at 18:45
















          Absolutely brilliant. Thanks a lot.
          – piotor
          Dec 8 at 18:45




          Absolutely brilliant. Thanks a lot.
          – piotor
          Dec 8 at 18:45


















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