1
$\begingroup$

Show that $\sum_{n=1}^{\infty} a^{\ln~n}$ is convergent if and only if $0<a<\frac{1}{e}$.

Proof:

$\ln~n<n$ for all $n \geq 1$.Hence the necessary condition for the series to be convergent will be $\lim_{n \rightarrow \infty} a^{\ln~n}=0$. And this will happen if $0<a<1$. Please give me some hint about how to proceed further.

$\endgroup$
1
  • $\begingroup$ As you've observed, one of the basic requirements for a series to converge is that the terms must go to zero...but this is not a very strong test, alas -- certainly not an if-and-only-if kind of thing. You'll need something more subtle. More important, though: your title asks about convergence of a power series, and yet this does not appear to be a power series. Did you mistype something? $\endgroup$ Oct 1, 2018 at 11:39

1 Answer 1

2
$\begingroup$

For $0<a<1$, $a^{\ln(n)}$ is a decreasing sequence. Hence by Cauchy condensation test, $$\sum a^{\ln(n)}\text{ is convergent} \iff \sum 2^na^{\ln(2^n)}\text{ is convergent} $$ The last sum is a geometric series, convergent for $0<a<1/e$.

Another approach would be to see that $a^{\ln(n)}=n^{\ln(a)}$, $\sum n^{\ln(a)}$ is a Dirichlet series, convergent only for $\ln(a)<-1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .