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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
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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.
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February 2nd, 2019, 12:43 PM   #3
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Quote:
Originally Posted by shashank dwivedi View Post
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.?
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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.
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February 9th, 2019, 01:23 PM   #5
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Thanks for rotation. Your analysis is correct.
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February 11th, 2019, 07:00 PM   #6
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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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