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

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3 Answers 3

up vote 9 down vote accepted

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

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

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

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and where i can see more about $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx?$$ –  slmkarta May 1 at 12:45
    
@slmkarta, If you mean examples, you have ample instances here –  lab bhattacharjee May 1 at 12:48
    
I mean the rules –  slmkarta May 1 at 12:50
    
@slmkarta, Put $b=1,a=0$ en.wikipedia.org/wiki/Riemann_sum#Methods –  lab bhattacharjee May 1 at 12:56

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

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