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 $f$.



By Riemann's sum


Refer also to the related

  • $\begingroup$ 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

Another approach is to consider that the same result is obtained by the series


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 ...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.