# Thread: problem with series limit..

1. ## problem with series limit..

$\displaystyle \lim_{n\to\infty}\sum_{n=1}^{\infty}\frac{1}{\sqrt {n^2+n}}=?$

i tried to look for collapsing fractions but they don't collapse

i tried to find the limit of the ratio of

$\displaystyle \frac{\sum_{n=1}^{\infty}\frac{1}{\sqrt{(n+1)^2+n+ 1}}}{\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+n}}}$

but i don't know how to begin untangling that thing

is there another way or can you tell me how to begin to figure something out of that?

2. ## Re: problem with series limit..

You have a variable scope problem in your problem statement. Once the sum has occurred, there is no n left in the expression. Thus, the limit basically does nothing. Did you mean to use a dummy variable for the sum, and sum, say, from j = 1 to n?

3. ## Re: problem with series limit..

Originally Posted by idom87
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+n}}=?$

Note that $\displaystyle \frac{1}{\sqrt{n^2+n}}\ge\frac{1}{\sqrt{2n^2}}$.

What can be said of $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{2}n}=?$

4. ## Re: problem with series limit..

i don't know, i can't decompose the fraction
what do you do in that case?

5. ## Re: problem with series limit..

Originally Posted by idom87
i don't know, i can't decompose the fraction
what do you do in that case?
First of all, is the problem as I changed correct?
It is a series and not a limit.

6. ## Re: problem with series limit..

it is correct and

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{2}n}\le\sum_{n=1 }^{\infty}\frac{1}{\sqrt{n^2+n}}$

7. ## Re: problem with series limit..

Originally Posted by idom87
it is correct and

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{2}~n}\le\sum_{n= 1}^{\infty}\frac{1}{\sqrt{n^2+n}}$
What can you say about the series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}~?$

8. ## Re: problem with series limit..

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{2}n}\le\sum_{n=1 }^{\infty}\frac{1}{\sqrt{n^2+n}}\le\sum_{n=1}^{ \infty }\frac{1}{n}$

i assume your'e refereing to the sandwich rule but i need to find the limits of the larger and smaller sums and i don't know how

9. ## Re: problem with series limit..

Originally Posted by idom87
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{2}n}\le\sum_{n=1 }^{\infty}\frac{1}{\sqrt{n^2+n}}\le\sum_{n=1}^{ \infty }\frac{1}{n}$
i assume your'e refereing to the sandwich rule but i need to find the limits of the larger and smaller sums and i don't know how
Why have you been asked to discuses series convergence when it is perfectly clear that you are missing the very basic ideas??

Has your text/instructor discussed the harmonic series?

If so what are the basics there?
How does that apply here?

If not, then you are not ready to do this question.

10. ## Re: problem with series limit..

ok i read about it in wikipedia and now i know and understand that $\displaystyle \sum_{n=1}^{ \infty }\frac{1}{n}\to\infty$

what next?
all of them $\displaystyle \to\infty$?

11. ## Re: problem with series limit..

Originally Posted by idom87
ok i read about it in wikipedia and now i know and understand that $\displaystyle \sum_{n=1}^{ \infty }\frac{1}{n}\to\infty$

what next?
all of them $\displaystyle \to\infty$?
Then any multiple of that also diverges.
$\displaystyle \sum {\frac{1}{{\sqrt 2 n}}}$ diverges.
Use basic comparison tests.

12. ## Re: problem with series limit..

so the answer is that the series diverges?

13. ## Re: problem with series limit..

Originally Posted by idom87
so the answer is that the series diverges?