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Can we find the limit of $$a_n=\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2}$$ using Riemann sum?

I can find the limit of $a_n$ using squeeze theorem and which is $0$. But can I apply Riemann sum to solve this problem? The question arise because the other problem $$b_n=\frac{1}{n}+\frac{1}{(n+1)}+\frac{1}{(n+2)}+\dots+\frac{1}{(2n)}$$ can be solved using Riemann sum as in this case we cannot apply squeeze theorem.

So I try, writing as, $$a_n= \frac{1}{n^2}+\frac{1}{n^2}\frac{1}{(1+\frac{1}{n})^2}+\frac{1}{n^2}\frac{1}{(1+\frac{2}{n})^2}+\dots+\frac{1}{(2n)^2}.$$ but this does not seem right to me. How can I do this? OR is it even doable using Riemann sum? If not then why not? Please help me here. Thanks.

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    $\begingroup$ Well $n\cdot a_n$ is a Riemann sum of $\int_0^1\frac{{\rm d}x}{(1+x)^2} = \frac{1}{2}$. $\endgroup$ – Winther Jun 5 '17 at 4:54
  • $\begingroup$ No need for that. $a_n$ is bounded by $\frac{n+1}{n^2}$ hence $\lim_{n\to +\infty}a_n = 0$, obviously. $\endgroup$ – Jack D'Aurizio Jun 5 '17 at 12:49
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EDIT: Sorry, I took a cursory glance at the problem and misunderstood so I included some stuff that doesn't matter. However, the latter part of my answer is pertinent.

For the first one, note that $$b_n = \sum^n_{i=0 }\frac{1}{n+i} = \frac{1}{n} \sum^n_{i=0} \frac{1}{1+\frac{i}{n}}.$$ This is exactly in the form of a Riemann sum, so we see $$\lim_{n\to \infty} b_n = \int_0^1 \frac{1}{1+x}dx = \ln(2).$$ For the second, I wouldn't suggest a Riemann sum, instead consider $$0 \le a_n = \sum^n_{i=0} \frac{1}{n^2(1 + i/n)^2} \le \sum^n_{i=0} \frac{1}{n^2} = \frac{n+1}{n^2} = \frac{1}{n} + \frac{1}{n^2}.$$ Thus taking the limit and using the squeeze theorem, we see $$0 \le \lim_{n\to \infty} a_n \le \lim_{n\to \infty} \left(\frac{1}{n} + \frac{1}{n^2}\right) = 0.$$

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  • $\begingroup$ I know that we can solve this by squeeze theorem. My question is, can we do it by Riemann sum or not? $\endgroup$ – Kushal Bhuyan Jun 5 '17 at 5:06
  • $\begingroup$ @KushalBhuyan Winther answered this 12 minutes before you posted this comment. Case solved? $\endgroup$ – Did Jun 5 '17 at 6:32
  • $\begingroup$ @Did yes indeed. Case solved now. $\endgroup$ – Kushal Bhuyan Jun 5 '17 at 15:14

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