My Math Forum Limit tending to infinity

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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.
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 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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Quote:
 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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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.
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 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

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