Evaluating the limit : $\displaystyle \lim_{n \to \infty} \dfrac{1}{\sqrt{n}} \displaystyle \sum_{k=1}^n \dfrac{1}{\sqrt{n+k}}$

This kind of question always baffles me. It looks like the answer is 0 but it isn't.

Can anyone tell me what does go on? And how do you evaluate this limit? $$\displaystyle \lim_{n \to \infty} \dfrac{1}{\sqrt{n}} \displaystyle \sum_{k=1}^n \dfrac{1}{\sqrt{n+k}}$$

Solution 1:

With or without taking a limit, there is a standard comparison between sum and integral of the same thing. Given real function $f(x)$ with $$ f > 0 \; , \; \; f' < 0 $$ and integers $a < b,$ $$ \int_a^{b+1} \; f(x) dx < \sum_{j=a}^b f(j) < \int_{a-1}^{b} \; f(x) dx$$

For you, $$ \int_{1}^{n+1} \; \frac{1}{\sqrt{x+n}} \; dx < \sum_{k=1}^n \frac{1}{\sqrt{k+n}} < \int_{0}^{n} \; \frac{1}{\sqrt{x+n}} \; dx$$

AFTERTHOUGHT: if you have a different problem with $f' >0$ both inequalities reverse, see Did I misuse the inductive hypothesis?

Solution 2:

Given

$\displaystyle \lim_{n \to \infty} \dfrac{1}{\sqrt{n}} \displaystyle \sum_{k=1}^n \dfrac{1}{\sqrt{n+k}} = \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum_{k=1}^n \dfrac{\sqrt{n}}{\sqrt{n+k}}=\displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum_{k=1}^n \dfrac{1}{\sqrt{1+\dfrac{k}{n}}}$

$\because \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum_{k=1}^n f\left(\dfrac{k}{n}\right) = \displaystyle \int_0^1 f(x) \mathrm dx$

$\implies \displaystyle \lim_{n \to \infty} \dfrac{1}{\sqrt{n}} \displaystyle \sum_{k=1}^n \dfrac{1}{\sqrt{n+k}} = \displaystyle \int_0^1 \dfrac{1}{\sqrt{1+x}} \mathrm dx = \displaystyle {\left. 2\sqrt{1+x}\right|_0^1}= 2\sqrt{1+1}-2\sqrt{1+0}= 2\sqrt2-2 $

$\boxed{\therefore\displaystyle \lim_{n \to \infty} \dfrac{1}{\sqrt{n}} \displaystyle \sum_{k=1}^n \dfrac{1}{\sqrt{n+k}}= \displaystyle{2\sqrt2-2}}$