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

• Hint: pull out a(nother) factor of $1/\sqrt{n}$ from the terms in the sum and see if you can recognize a Riemann sum in the result. – Steven Stadnicki Oct 5 '18 at 17:30
• The sum is equal to $$\frac1n \sum_{k=1}^n \frac{1}{\sqrt{1+ \frac kn}} \to \int_0^1 (1+t)^{-\frac12} dt = 2(\sqrt{2} -1).$$ – nguyen0610 Oct 5 '18 at 17:31
• Also, note that the sum is bounded from below by $\sum_{k=1}^n1/\sqrt{n+n}$ $=(1/\sqrt{2})\sum_{k=1}^n1/\sqrt{n}$ $=(1/\sqrt{2})(n\times 1/\sqrt{n})$ $=\sqrt{n}/\sqrt{2}$, so your limit certainly can't be zero. – Steven Stadnicki Oct 5 '18 at 17:33
• @user599663: this is tagged riemann-sum. – robjohn Oct 5 '18 at 17:35
• Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn Oct 5 '18 at 17:43

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?