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Hi i need to calculate limit of integral sequence: $$\lim_{n\rightarrow\infty}\frac{1}{n}\int_{1}^{\infty}\frac{dx}{x^2\cdot\ln{(1+\frac{x}{n})}}=\lim_{n\rightarrow\infty}\int_{1}^{\infty}\frac{dx}{x^2\cdot n\cdot\ln{(1+\frac{x}{n})}}=\int_{1}^{\infty}\frac{dx}{x^2\cdot e^x}$$ and it's true if i can use monotone convergence theorem. Everything seem fine but i am not sure about monotonicity of $f_n$. I thinks it's decreasing and according to statement it should be otherwise. And taking negative doesn't seem right at all. And another one. $$\lim_{n\rightarrow\infty}\int_{0}^{n}(1+\frac{x}{n})^{n+1}\cdot e^{-2x}=\int_{0}^{\infty}e^{-x}$$ and here i am not sure about the $\infty$ in integral. I am hopeless newbie so i will be glad for thorough explanation.

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    $\begingroup$ You have computed the wrong limit. You could a) use the dominated convergence theorem, or b) use a variant of the monotonone convergence theorem; you have $f_n \geqslant f_{n+1} \geqslant 0$, that suffices to interchange limit and integral, since $f_1$ is integrable. You could also look at $g_n = f_1 - f_n$ and apply the MCT to that. $\endgroup$ – Daniel Fischer Jan 17 '16 at 12:19
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If you are not familiar with dominated convergence and such things, you could use the inequality $$ \frac{y}{1+y}<\ln(1+y)<y $$ which yields $$ \int_1^{+\infty}\frac{1}{x^3}\,dx<\frac{1}{n}\int_1^{+\infty}\frac{1}{x^2\ln(1+x/n)}\,dx<\int_1^{+\infty}\frac{1}{x^3}+\frac{1}{n}\frac{1}{x^2}\,dx. $$ Calculate the integrals and use the squeeze theorem, and you will see that the limit is $1/2$.

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Change variables in your first integral $x/n=t$, yielding $$ \lim_{n\to\infty}\frac{1}{n}\frac{n}{n^2}\int_{1/n}^\infty \frac{dt}{t^2\ln(1+t)}=\lim_{n\to\infty}\frac{1}{n^2}\int_{1/n}^\infty \frac{dt}{t^2\ln(1+t)}\ . $$ Then use L'Hopital and compute $$ \lim_{n\to\infty}\frac{\partial_n \int_{1/n}^\infty \frac{dt}{t^2\ln(1+t)}}{2n}\ , $$ whose numerator can be evaluated using the fundamental theorem of calculus $$ \partial_n \int_{1/n}^\infty \frac{dt}{t^2\ln(1+t)}=\frac{1}{n^2}\frac{1}{(1/n)^2\ln (1+1/n)}\ . $$ Eventually, your limit is equivalent to computing $$ \lim_{n\to\infty}\frac{1}{2n\ln(1+1/n)}=1/2\ . $$

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Note that, for $x\ge1$, $$ \lim_{n\to\infty}\frac{1}{x^2}\frac{1}{n \ln(1+x/n)}=\frac{1}{x^2}\frac{1}{\ln(e^x)}=\frac{1}{x^3}. $$ Furthermore such an integrand is controlled as follows $$ \frac{1}{x^2}\frac{1}{n \ln(1+x/n)}\le \frac{1}{x^2}\frac{1}{n \ln(1+1/n)} $$ and since $\lim[n\ln(1+1/n)]=1$, there will be a constant $C$ such that $$ \frac{1}{x^2}\frac{1}{n \ln(1+x/n)}\le\frac{C}{x^2}; $$ in fact, it can be shown that $(1+1/n)^n$ is monotone and increasing, so that $C=1/\ln2$ is fine, by the monotonicity of the logarithm. Since the integrand has a ''guardian'' $C/x^2$ (an estimate from above) which is summable, $$ \int_{1}^{+\infty}\frac{C}{x^2}=C, $$ we can apply Lebesgue's dominated convergence theorem and take the limit $n\to\infty$ inside the integral sign: $$ \lim_{n\to\infty}\int_{1}^{+\infty}\frac{1}{x^2n\ln(1+x/n)}dx= \int_{1}^{+\infty}\lim_{n\to\infty}\frac{1}{x^2n\ln(1+x/n)}dx=\int_0^{+\infty}\frac{1}{x^3}dx=1/2. $$ Similarly, you can work out the second example.

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  • $\begingroup$ Alright i get it know. Thanks for solid and detailed explanation. $$\int_{0}^{n}(1+\frac{x}{n})^{n+1}\cdot e^{-2x}dx=\int_{0}^{n}(1+\frac{x}{n})^{n}\cdot (1+\frac{x}{n})\cdot e^{-2x}dx\le\int_{0}^{n}C\cdot e^{-x}dx\le\int_{0}^{\infty}C\cdot e^{-x}dx$$ as exponential function is monotone and increasing it should be true. Bo unlike the previous $C$ may be arbitrary big(we take $x$ as fixed value?) or my esestimation is wrong. And what about $n$ in integral? I can freely go with it to infinity? $\endgroup$ – vanHohenheim Jan 17 '16 at 19:44
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    $\begingroup$ @vanHohenheim The $C$-estimate is fine, since $(1+x/n)^n\to e^x$ from below. The last step is also correct because the integral of a positive function on a larger domain is always bigger than its integral on a smaller one! Note that you can always think of $\int_0^nf(x)dx$ as $\int_0^{+\infty}\chi_{[0,n]}(x)f(x)dx$, where $\chi$ is the characteristic function, so all in all there is nothing exoteric in sending $n\to\infty$ in the integration bound. $\endgroup$ – Brightsun Jan 17 '16 at 20:10

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