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February 2nd, 2019, 05:12 AM   #1
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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. Attached Images 20190202_193438.jpg (98.0 KB, 35 views) 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
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 Originally Posted by shashank dwivedi 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. Could you rotate picture 90 deg.? February 8th, 2019, 04:43 PM   #4
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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.
Attached Images limit_2019_02_08_rotated.jpg (49.8 KB, 10 views) February 9th, 2019, 01:23 PM #5 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Thanks for rotation. Your analysis is correct. Thanks from jks 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, $\displaystyle \lim_{n\to\infty}\int_0^1\frac{1}{\sqrt{1+t/n^2}}\,dt\implies\int_0^1\left(\lim_{n\to\infty} \frac{1}{\sqrt{1+t/n^2}}\right)\,dt=\int_0^1\,dt=1$ Thanks from jks, idontknow and shashank dwivedi Tags infinity, limit, tending Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Katya Calculus 1 November 16th, 2016 02:38 AM cacophonyjm Calculus 8 March 11th, 2012 07:29 PM joefrank87 Advanced Statistics 2 July 17th, 2011 01:51 PM mechintosh Real Analysis 1 March 1st, 2010 06:54 PM rajend3 Calculus 6 August 21st, 2009 10:17 AM

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