February 2nd, 2019, 05:12 AM  #1 
Member Joined: Apr 2017 From: India Posts: 73 Thanks: 0  Limit tending to infinity
I am unable to find out the limit of this entire sum with n tending to infinity. I know the concept of definite integral. However, I am unable to apply that. Please help me out. 
February 2nd, 2019, 10:05 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 600 Thanks: 87 
The limit is not 0, as you selected it. I'm not sure, but the limit must be 1. Last edited by skipjack; February 8th, 2019 at 07:34 PM. 
February 2nd, 2019, 12:43 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716  
February 8th, 2019, 04:43 PM  #4 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications 
I have attached a file with the image rotated. I think that the Squeeze Theorem may be used. As stated in the link, we bound a function $f(n)$ on both sides: $\displaystyle g(n) \leq f(n) \leq h(n)$ Specific to this problem: $\displaystyle f(n)=\sum_{i=1}^n \frac{1}{\sqrt{n^2+i}}$ and we are to find: $$\lim_{n\to\infty} f(n)$$ Let us choose a function $h(n)$ for an upper bound: $\displaystyle h(n)=\sum_{i=1}^n \frac{1}{\sqrt{n^2}}=\frac{n}{n}=1$ Note that every term in $f(n)$ is less than every term in $h(n)$, hence $h(n)$ is an upper bound. Obviously: $$\lim_{n\to\infty} h(n)=1$$ Let us choose a function $g(n)$ for a lower bound: $\displaystyle g(n)=\sum_{i=1}^n \frac{1}{\sqrt{n^2+n+\frac{1}{4}}}=\frac{n}{n+ \frac{1}{2}}$ Note that every term in $g(n)$ is less than every term in $f(n)$, hence $g(n)$ is a lower bound. And: $$\lim_{n\to\infty} g(n)=1$$ Thus: $$\lim_{n\to\infty} f(n)=1$$ I hope that I did this correctly. 
February 9th, 2019, 01:23 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716 
Thanks for rotation. Your analysis is correct.

February 11th, 2019, 07:00 PM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 
We have $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{n^2+k} }$ By the Limit Definition of the Definite integral $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{n^2+k} }=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{1+k/n^2}}\frac1n\implies\lim_{n\to\infty}\int_0^1\frac {1}{\sqrt{1+t/n^2}}\,dt$ By the Monotone convergence theorem, $\displaystyle \lim_{n\to\infty}\int_0^1\frac{1}{\sqrt{1+t/n^2}}\,dt\implies\int_0^1\left(\lim_{n\to\infty} \frac{1}{\sqrt{1+t/n^2}}\right)\,dt=\int_0^1\,dt=1$ 

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infinity, limit, tending 
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