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What would be the sum of following ?

$$\lim_{n\to\infty} \left[\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + \cdots + \frac{1}{(n+n)^{2}}\right]$$

I tried to turn it into integral :

$\displaystyle\int \frac{1}{(1+\frac{r}{n})^{2}}\frac{1}{n^{2}} $ but I can't figure out how to deal with $\frac{1}{n^{2}}$

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For large values it will be $0$ i think – Archis Welankar Mar 1 at 14:26
The sum is asymptotic to $\log 2/n$, isn't it? – Nima Bavari Mar 1 at 21:31
I really don't know the final answer. – Mojo Jojo Mar 2 at 5:56
up vote 8 down vote accepted

Hint: $\frac{1}{(n+k)^2}\leq \frac{1}{n^2}$ so what is an upper bound for:


By the way, you could see this as an integral limit if you wrote it as:

$$\frac{1}{n}\sum_{k=1}^{n} \frac{1}{n}\frac{1}{\left(1+\frac k n\right)^2}$$

But: $$\sum_{k=1}^{n} \frac{1}{n}\frac{1}{\left(1+\frac k n\right)^2}\to\int_{0}^1 \frac{dx}{(1+x)^2}=\frac12$$

So with the extra $\frac{1}{n}$, you can determine what the limit will be...

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Thanks, didn't think this approach. – Mojo Jojo Mar 1 at 14:52
excuse me but isnt this supposed to be $\int_{1}^2 \frac{dx}{(1+x)^2}$ ? – Agawa001 Mar 2 at 11:42
Whoops, yes, will fix when I get to a computer. @Agawa001 – Thomas Andrews Mar 2 at 12:20
Also, the range was wrong - should be 0 to 1. – Thomas Andrews Mar 2 at 12:23
mm yes i used to be much focused on the container than bounds of integral. but what you did is nice trick – Agawa001 Mar 2 at 14:20

$$\lim_{n\to\infty}0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le\lim_{n\to\infty}\frac{n}{(n+1)^2}$$ $$\lim_{n\to\infty}0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le\lim_{n\to\infty}\frac{\frac1n}{(1+\frac1n)^2}$$ $$0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le0$$ $$\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]=0$$

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The series $\sum_{j=}^{\infty}1/j^2$ converges by the Cauchy condensation test so $0<\sum_{j=n}^{j=2 n}1/j^2 < \sum_{j=n}^{\infty}1/j^2$ which $\to 0$ as $n\to \infty.$

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