As shown in the title, I'm evaluating the following:$\lim_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^{2n}\frac{1-\ln(1+\frac{k}{n})}{(1+\frac{k}{n})^2}$ And I get stuck. Any ideas are welcome.
Evaluate $\lim_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^{2n}\frac{1-\ln(1+\frac{k}{n})}{(1+\frac{k}{n})^2}$
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algebra-precalculus
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0@m0nhawk That will overcomplicate things. – 2012-11-14
2 Answers
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Consider funstion $ f(x)=\frac{1-\log (1+x)}{(1+x)^2} $ and partition $\{k/n:k=\overline{1,2n}\}$ of the interval $[0,2]$. Then Riemann sum for given partition will be... Then recall Riemann sums tends to integral when partition becomes smaller, i.e. when $1/n\to 0$.
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Consider the partitions
$\left\{0<\frac{1}{n}<\frac{2}{n}<\ldots <\frac{2n}{n}=2\right\}\,\,\text{of the interval}\,[0,2]\,\,,\,\,n\in\Bbb N$
Choosing the right-end points of each interval, we get:
$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1-\log\left(1+\frac{k}{n}\right)}{\left(1+\frac{k}{n}\right)^2}=\int_0^2\frac{1-\log(1+x)}{(1+x)^2}\,dx=$
$=\left.-\frac{\log(x+1)+1}{x+1}\right|_0^2=\ldots$
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1Yes, of course...but you know $\int\frac{1}{x^2}dx=-\frac{1}{x}$right? – 2012-11-18