1. ## Infinite Series

I need some serious help in the attached questions...I need to determine whether the series in the picture converge, absolutely converge ot diverge...
I realy need your guidance and help....

Thanks everyone

My attemp:

I've no clue aboue 1&3...
In 2-I know that cos(n*pi)*ln([n+1]/n ) converges by leibnitz test...But what about the first element in the sum? [sqrt(n)sin(n)tan(1/n^2)] ...

2. Originally Posted by WannaBe
I need some serious help in the attached questions...I need to determine whether the series in the picture converge, absolutely converge ot diverge...
I realy need your guidance and help....

Thanks everyone

My attemp:

I've no clue aboue 1&3...
In 2-I know that cos(n*pi)*ln([n+1]/n ) converges by leibnitz test...But what about the first element in the sum? [sqrt(n)sin(n)tan(1/n^2)] ...
What have you done?

$\displaystyle 1-n\sin\left(\frac{1}{n}\right)1-n\left(\frac{1}{n}-\frac{1}{6n^3}+\cdots\right)=\frac{1}{6n^2}+\cdots \underset{n\to\infty}{\sim}\frac{1}{6n^2}$. So this sum converges.

3. The third is equally uneventful combine the fractions note the behavior of the numerator and denominator and seperately

Spoiler:

$\displaystyle \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}\underset{n\to\infty}{\sim}\fr ac{1}{n^{\frac{3}{2}}}$

4. For the third one. I WILL not give you the answer. Consider this though $\displaystyle \lim_{z\to0}\frac{\tan(z)}{z}$

5. Tnx a lot ! ...

6. Originally Posted by Drexel28
The third is equally uneventful combine the fractions note the behavior of the numerator and denominator and seperately

$\displaystyle \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}\underset{n\to\infty}{\sim}\fr ac{1}{n^{\frac{3}{2}}}$
The general term of the third series seems to be...

$\displaystyle \frac{1}{\sqrt{n}-1}-\frac{1}{\sqrt{n}+1}\underset{n\to\infty}{\sim}\fr ac{2}{n}$

... so that the series diverges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

7. Originally Posted by chisigma
The general term of the third series seems to be...

$\displaystyle \frac{1}{\sqrt{n}-1}-\frac{1}{\sqrt{n}+1}\underset{n\to\infty}{\sim}\fr ac{2}{n}$

... so that the series diverges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
ITS
$\displaystyle \frac{1}{\sqrt{n+1}-1}-\frac{1}{\sqrt{n+1}+1}$
Assuimg n > 1