sequences-and-series - self learning - necessary and sufficient condition for convergence of power series - answerstu - answerstu

1 Answer

  1. John- Reply


    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 $\ln(a)<-1$.

Leave a Reply

Your email address will not be published. Required fields are marked *

You can use these HTML tags and attributes <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>