Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

given the series

$$ S=\sum_{n=1}^{\infty}n^{a} $$ for $ Re(a) <1 $ is this series Cesaro summable of any finite order ??

share|cite|improve this question
If $\Re(a)<-1$, zeroeth order works. – Hagen von Eitzen Feb 15 '13 at 21:31
up vote 2 down vote accepted

Let $x^a=(x^a_n)_n$ defined by $x^a_n=n^a$. For every $(x_n)_n$, call $(C(x)_n)_n$ the sequence of its Cesàro's sums, defined by $C(x)_n=\frac1n\sum\limits_{k=1}^nx_k$.

If $\Re(a)\lt-1$, $x^a$ is absolutely summable.

If $\Re(a)\gt-1$, comparing $C(x^a)_n$ to a Riemann sum, one sees that $$ C(x^a)_n=n^a\frac1n\sum_{k=1}^n\left(\frac{k}n\right)^a\sim n^a\int_0^1t^a\mathrm dt=\frac{x^a_n}{a+1}. $$ Thus, for every $k\geqslant1$, $C^{(k)}(x^a)_n\sim\frac1{(a+1)^k}x^a_n$, hence no $C^{(k)}(x^a)$ is summable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.