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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0The last substitution doesn't make sence. Previous exrcise was given to simplify evaluation of integrals you will face in while calulating this sum – 2012-11-14
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0Think about Riemann sums – 2012-11-14
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0Have you tried expanding $log(1+x) = x - x^2/2 + \ldots$? I think this can be useful, because the $k/n < 1$. – 2012-11-14
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0It turns into sum of infinite sum, even troublesome. – 2012-11-14
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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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0The symbol between the limit and the integral must be an "=". Unless that, everything is fine. – 2012-11-14
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0Of course, thanks – 2012-11-14
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0The last step is weird, how can you reach it? – 2012-11-18
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1What do you mean "weird"? That's the integral's primitive, of course: putting $$u=\log x\;\;,\;u'=\frac{1}{x}\;;\;v'=\frac{1}{x^2}\;\;,\;v=-\frac{1}{x}$$one gets at once integrating by parts: $$\int\frac{\log x\,dx}{x^2}=-\frac{\log x+1}{x}$$Now pass from $\,x\,$ to $\,x+1\,$...voila! – 2012-11-18
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0But the integral is:$$\int \frac{1-log (x)}{x^2} dx$$ – 2012-11-18
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1Yes, of course...but you know $$\int\frac{1}{x^2}dx=-\frac{1}{x}$$right? – 2012-11-18