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Show that $$\lim_{n\to\infty}n\Bigg\{\dfrac{1}{(n+1)^2}+\dfrac{1}{(n+2)^2}+\dfrac{1}{(n+3)^2}+\cdots+\dfrac{1}{(2n)^2}\Bigg\}=\dfrac{1}{2}.$$

Proof: We can rewrite $$\lim_{n\to\infty}n\Bigg\{\dfrac{1}{(n+1)^2}+\dfrac{1}{(n+2)^2}+\dfrac{1}{(n+3)^2}+\cdots+\dfrac{1}{(2n)^2}\Bigg\}=\lim_{n\to\infty}n\Bigg\{\sum_{k=1}^n\dfrac{1}{(n+k)^2}\Bigg\}$$ Which looks astoundingly similar to the form in Corollary 8.3, where $\dfrac{b-a}{n} = n\implies b-a = 2n$.

Corollary 8.3: Let $f$ be a continuous function on an interval [a,b]. Then $$\int_a^bf(x)dx=\lim_{n\to\infty}\dfrac{b-a}{n}\sum_{k=0}^{n-1}f\Bigg(a+\dfrac{k}{n}(b-a)\Bigg)$$

That would mean the term in Corollary 8.3 would morph into $$f\left(a+\dfrac{k}{n}(b-a)\right)=f\left((b-2n)+\dfrac{k}{n}(2n)\right)=f(b-2n+2k)$$ We need $f(b-2n+2k)$ to look like $\dfrac{1}{(n+k)^2}$. This would imply that $f$ is

... but then I get stuck. I am unsure how to find $f$ or even if I need to find $f$.

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HINT

By Riemann's sum

$$\lim_{n\to\infty}n\sum_{k=1}^n\dfrac{1}{(n+k)^2}=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\dfrac{1}{(1+k/n)^2}$$

Refer also to the related

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  • $\begingroup$ But then I get to here: $\lim_{n\to\infty}\dfrac{n}{n}\Bigg\{\sum_{k=1}^n\dfrac{1}{(1+(k/n))^2}\Bigg\}=\lim_{n\to\infty}\Bigg\{\sum_{k=1}^n\dfrac{1}{(1+(k/n))^2}\Bigg\}$, which is infinity. EDIT: Ah I see. I forgot to square. $\endgroup$ – kaisa Dec 5 '18 at 11:09
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Another approach is to consider that the same result is obtained by the series

$$\lim_{n\to\infty}{\sum_{k=1}^n\dfrac{n}{(n+k)(n+k+1)}}$$

Proof: the difference of the two terms is $O(n^{-2})$, $O(n^{-1})$ after the summation.

And then to use $$\frac{1}{(n+k)(n+k+1)} = \frac{1}{n+k} - \frac{1}{n+k+1} $$

Which simplifies the summation ...

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