Prove that if $a_k≥0$ and $\Sigma \, a_k$ is $(C,1)-summable$ (Cesáro summable), then the series is convergent in the usual sense. (Assume the contrary – what does that entail for a positive series?)
I can't really get the hang of this, I am in the beginning of my Fourier analysis course and we are laying the groundwork at the moment. I really want to understand this because I am thinking about taking a master in pure mathematics. Anyway, this is what I am thinking
Since $\Sigma \, a_k$ is $(C,1)$-summable we know that the value of
$$ \sigma_n=\frac{S_1+S_2+...+S_n}{n} $$ has a finit value as $n$ tends to infinity.
Can I use this for something?
Since $a_k \geqslant 0$, the sequence
$$S_n = \sum_{k = 1}^n a_k$$
of partial sums is increasing (non-strictly in general). So it is convergent (to a real number) if and only if it is bounded.
We also know that if $(S_n)$ converges to $L\in \mathbb{R}$, then the sequence $(\sigma_n)$ of Cesàro means of $(S_n)$ also converges to $L$. So if the assertion we want to prove is true, then we can bound $(S_n)$ by $S = \lim\limits_{n\to\infty} \sigma_n$. But we need not prove that sharp bound, any bound would do, and less sharp bounds might be easier to prove (here, there's no real difference; once you know the idea, all bounds are proved in essentially the same way).
The idea is that since $(S_n)$ is monotonically increasing, we can get a lower bound for $\sigma_m$ in terms of $S_n$ when $n \leqslant m$ by ignoring the $S_k$ for $k < n$ and replacing the $S_k$ for $k > n$ with $S_n$:
\begin{align} \sigma_m &= \frac{1}{m}\sum_{k = 1}^m S_k\\ &\geqslant \frac{1}{m} \sum_{k = n}^m S_k\\ &\geqslant \frac{1}{m} \sum_{k = n}^m S_n\\ &= \frac{m - n + 1}{m} \cdot S_n. \end{align}
Rearranging yields
$$S_n \leqslant \frac{m}{m-n+1}\cdot \sigma_m\tag{1}$$
for all $m \geqslant n$. Letting $m \to \infty$ in $(1)$, we obtain the sharp bound $S_n \leqslant S$ since $\lim\limits_{m\to\infty} \frac{m}{m-n+1} = 1$ and $\sigma_m \to S$.
If we take $m = 2n$ in $(1)$, we get the bound $S_n \leqslant 2 \sigma_{2n}$, which together with $\sigma_n \leqslant S_n$ shows that $(S_n)$ is bounded if and only if $(\sigma_n)$ is bounded without assuming that $(\sigma_n)$ converges, so either both converge or neither does.