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

$$\lim_{n\to\infty}\bigg(\frac{1}{\sqrt{9n^2-1^2}}+\frac{1}{\sqrt{9n^2-2^2}}+ \dots +\frac{1}{\sqrt{9n^2-n^2}}\bigg)$$ I need a hint. I see that maybe compute with integral. But what the integrable function?

share|cite|improve this question
up vote 10 down vote accepted

Hint Since you ask for a simple hint then: Use a Riemann sum and compute this integral $$\int_0^1\frac{dx}{\sqrt{9-x^2}}$$

share|cite|improve this answer

$$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

Here the $r$( where $1\le r\le n$)th term $=\displaystyle\frac n{\sqrt{9n^2-r^2}}=\frac1{\sqrt{9-\left(\frac rn\right)^2}}$

share|cite|improve this answer
and where i can see more about $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx?$$ – Simankov May 1 '14 at 12:45
@slmkarta, If you mean examples, you have ample instances here – lab bhattacharjee May 1 '14 at 12:48
I mean the rules – Simankov May 1 '14 at 12:50
@slmkarta, Put $b=1,a=0$ – lab bhattacharjee May 1 '14 at 12:56

Well, here is a hint, factor out $n$, and use it to convert it into Riemann Integration.

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.