Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Without using integrals, how to find this limit:

$$\mathop {\lim }\limits_{n \to \infty } {a_n} = n\cdot\left({1 \over {{{(n + 1)}^2}}} + {1 \over {{{(n + 2)}^2}}} + \cdots{1 \over {{{(2n)}^2}}}\right)$$

I tried squeezing the sequence but it didn't workout.
What next should I do?

share|cite|improve this question
The tag (limit-theorems) is not a good fit for this questions, see the tag-wiki. – Martin Sleziak Nov 29 '13 at 8:05
up vote 5 down vote accepted

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

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

Now use squeeze theorem to find $\lim_{n \to \infty}{a_n}$.

share|cite|improve this answer
telescoping series. nice one! – Daniel Gagnon Nov 27 '13 at 20:54

It's value is equal to $\dfrac{1}{2}$, using definite integral as a limit of sum where $f(x)=\frac{1}{x^2}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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