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given the series

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

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If $\Re(a)<-1$, zeroeth order works. –  Hagen von Eitzen Feb 15 '13 at 21:31

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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.

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