How does Cesaro summability imply the partial sums converge to the same sum?
Olivia Zamora 
I can't reconcile this fact I used to know.
Suppose you have a sequence of nonnegative terms $a_k$. Let $s_n=\sum_{k=1}^n a_k$, and suppose $$ \lim_{n\to\infty}\frac{s_1+\cdots+s_n}{n}=L. $$ Then $\sum_{k=1}^\infty a_k$ also exists and equals $L$.
I could recover that $\sum_{k=1}^\infty a_k$ exists. If not, it diverges to $\infty$. Suppose $M>0$ is given. There exists $N$ such that $\sum_{k=1}^N a_k>M$. If $n>N$, then $$ \begin{align*} \frac{s_1+\cdots+s_n}{n} &=\frac{s_1+\cdots+s_N}{n}+\frac{s_{N+1}+\cdots+s_n}{n}\\ &\geq\frac{n-N}{n}M. \end{align*} $$ Taking $n\to\infty$ shows that $L\geq M$ for all positive $M$, which is clearly not true.
But I can't for the life of me remember why $\sum_{k=1}^\infty a_k=L$ and can't find it online. Can someone clear this up for me? Thanks.
$\endgroup$ 22 Answers
$\begingroup$You already showed that $\lim_{n \to \infty} s_n = L'$ exists.
Assume now that $L' < L$. Then $s_n \le L'$ for all $n$, since the $a_n$ are non-negative. Therefore $ \frac{s_1 + s_2 + \dots + s_n}{n} \le L'$ for all $n$, which contradicts the assumption.
Assume on the other hand that $L' > L$. Then $s_n > \frac{L' + L}{2}$ for sufficiently large $n$, say $n \ge K$. Therefore
$$
\frac{s_1 + s_2 + \dots + s_n}{n} = \frac{s_1 + s_2 + \dots + s_K}{n} + \frac{s_{K+1} + \dots + s_n}{n} \ge \frac{(n-K)(L'+L)}{2n}
$$
and as $n \to \infty$ you get another contradiction.
No this is not true. $a_k = (-1)^k$ is a counterexample.
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