1
$\begingroup$

If we have $(n^2+a^2)$ instead of $(n^2+a)$ in the denominator, we just evaluate ${\int_{0}^{1}{\frac1{\sqrt{1+x^2}}dx}}$. But how do I express this above form as a definite integral?

$\endgroup$
1
  • 2
    $\begingroup$ I wouldn't express it as an integral. A simple estimate yields the limit. $\endgroup$ Apr 30, 2014 at 15:41

2 Answers 2

4
$\begingroup$

HINT:

$$n\cdot\frac1{\sqrt{n^2+n}}<\frac1{\sqrt {n^2+1}}+\frac1{\sqrt {n^2+2}}+\cdots+\frac1{\sqrt {n^2+n}}< n\cdot\frac1{\sqrt{n^2}}$$

$\endgroup$
1
  • $\begingroup$ Thanks, got it. I absolutely forgot about this approach! $\endgroup$
    – Rajarshi
    Apr 30, 2014 at 15:48
1
$\begingroup$

You want to calculate

$$X = \lim_{n\to\infty} \left( \sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} \right)$$

You can estimate it from both sides:

$$\lim_{n\to\infty} \left( \sum_{k=1}^n \frac{1}{\sqrt{n^2}} \right) \leq X \leq \lim_{n\to\infty} \left( \sum_{k=1}^n \frac{1}{\sqrt{n^2+n}} \right)$$

This can be simplified to

$$1 = \lim_{n\to\infty} \left( \frac{n}{\sqrt{n^2}} \right) \leq X \leq \lim_{n\to\infty} \left( \frac{n}{\sqrt{n^2+n}} \right) = 1$$

so, the solution is $X=1$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .