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.

Attached Images

20190202_193438.jpg (98.0 KB, 35 views)

idontknow

The limit is not 0, as you selected it.
I'm not sure, but the limit must be 1.

mathman

Could you rotate picture 90 deg.?

jks

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)

mathman

Thanks for rotation. Your analysis is correct.

greg1313

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,

$\displaystyle \lim_{n\to\infty}\int_0^1\frac{1}{\sqrt{1+t/n^2}}\,dt\implies\int_0^1\left(\lim_{n\to\infty} \frac{1}{\sqrt{1+t/n^2}}\right)\,dt=\int_0^1\,dt=1$