Proving Convergence of a Series with Parameter a

[twoflower]

Hi,

I can't find a way to prove the convergence of the following sum regarding to parameter a:

<br /> \sum_{n=1}^{ \infty } a_{n}<br />

where

<br /> a_{n} = \frac { \sqrt{1+ \frac{1}{n}} - 1}{n^{a}}<br />

I already proved the necessary condition for convergence, ie that

<br /> \lim_{n \rightarrow \infty} a_{n} = 0<br />

And it showed that a must be in (0, \infty).

But I can't figure out how to prove the convergence. I tried d'Alembert's criterion, comparing criterion, limite comparing criterion but no gave me some useful result (with d'Alembert I got very complicated expression I wasn't able to simplify).

And one more question: in school I didn't understand, whether there is a equivalency in Abel-Dirichlet's criterion for convergence. I mean if neither condition of the theorem is passed, could we say that the sum diverges?

Thank you.

 

[arildno]

Well, but:
a_{n}=\frac{1}{(1+\sqrt{1+\frac{1}{n}})n^{(a+1)}}\leq\frac{1}{n^{(a+1)}}

Hence, each a_{n} is bounded by a term in a convergent series.

 

[shmoe]

And one more question: in school I didn't understand, whether there is a equivalency in Abel-Dirichlet's criterion for convergence. I mean if neither condition of the theorem is passed, could we say that the sum diverges?

To apply Abel-Dirichlet's criterea you have to break the terms of your series a_n, into a product a_n=b_{n}c_{n}. Depending on how you choose the b and c sequences will affect whether the conditions are satisfied. If you find one such decomposition that works, you have convergence. If you find one that fails, it doesn't mean that they will all fail.

 

[twoflower]

Well, but:
a_{n}=\frac{1}{(1+\sqrt{1+\frac{1}{n}})n^{(a+1)}}\leq\frac{1}{n^{(a+1)}}

Hence, each a_{n} is bounded by a term in a convergent series.

Clear and simple.. Thank you very much arildno.

Btw isn't there any list of most common series types and corresponding criterion, which is most usable in the situation?

 

[twoflower]

To apply Abel-Dirichlet's criterea you have to break the terms of your series a_n, into a product a_n=b_{n}c_{n}. Depending on how you choose the b and c sequences will affect whether the conditions are satisfied. If you find one such decomposition that works, you have convergence. If you find one that fails, it doesn't mean that they will all fail.

Thank you shmoe, that's exactly what I was asking for.

 

[arildno]

Clear and simple.. Thank you very much arildno.

Btw isn't there any list of most common series types and corresponding criterion, which is most usable in the situation?

If there is, it's not inside my head..