twoflower
Dec3-04, 11:04 AM
Hi,
I can't find a way to prove the convergence of the following sum regarding to parameter a:
\sum_{n=1}^{ \infty } a_{n}
where
a_{n} = \frac { \sqrt{1+ \frac{1}{n}} - 1}{n^{a}}
I already proved the neccessary condition for convergence, ie that
\lim_{n \rightarrow \infty} a_{n} = 0
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.
I can't find a way to prove the convergence of the following sum regarding to parameter a:
\sum_{n=1}^{ \infty } a_{n}
where
a_{n} = \frac { \sqrt{1+ \frac{1}{n}} - 1}{n^{a}}
I already proved the neccessary condition for convergence, ie that
\lim_{n \rightarrow \infty} a_{n} = 0
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.