Real Analysis, $limlimits_{nrightarrowinfty} int_{0}^{1} frac{e^{-nt}-(1-t)^n}{t} dt$
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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
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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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
edited Dec 8 '18 at 11:46
Batominovski
33k33293
33k33293
asked Dec 8 '18 at 10:24
piotorpiotor
405
405
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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 $n=1,2,3,ldots$.
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Absolutely brilliant. Thanks a lot.
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– piotor
Dec 8 '18 at 18:45
add a comment |
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1 Answer
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$begingroup$
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 $n=1,2,3,ldots$.
$endgroup$
$begingroup$
Absolutely brilliant. Thanks a lot.
$endgroup$
– piotor
Dec 8 '18 at 18:45
add a comment |
$begingroup$
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 $n=1,2,3,ldots$.
$endgroup$
$begingroup$
Absolutely brilliant. Thanks a lot.
$endgroup$
– piotor
Dec 8 '18 at 18:45
add a comment |
$begingroup$
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 $n=1,2,3,ldots$.
$endgroup$
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 $n=1,2,3,ldots$.
answered Dec 8 '18 at 11:39
BatominovskiBatominovski
33k33293
33k33293
$begingroup$
Absolutely brilliant. Thanks a lot.
$endgroup$
– piotor
Dec 8 '18 at 18:45
add a comment |
$begingroup$
Absolutely brilliant. Thanks a lot.
$endgroup$
– piotor
Dec 8 '18 at 18:45
$begingroup$
Absolutely brilliant. Thanks a lot.
$endgroup$
– piotor
Dec 8 '18 at 18:45
$begingroup$
Absolutely brilliant. Thanks a lot.
$endgroup$
– piotor
Dec 8 '18 at 18:45
add a comment |
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