I've been studying for an analysis qual and this problem was on one of the past exams.

Let $\{a_{j}\}$ $\subseteq$ $\mathbb{R}$ such that $\sum_{j=1}^{\infty}a_{j}=\frac{3\pi}{4}$. For every $n\in \mathbb{N}$, define $T_{n}=\frac{1}{n}\sum_{j=1}^{n}S_{j}$ where for each $j \in \mathbb{N}$, $S_{j}=\sum_{k=1}^{j} a_{k}$.

Does $\{ T_{n} \}_{n=1}^{\infty}$ converge? If so, find its sum.

I have been able to show that |$T_{n}$| is bounded by $\frac{3\pi}{4}$, but I have a sneaking suspicion that it does indeed converge to $\frac{3\pi}{4}$ but have been unable to show it. Can I get some help with that?


  • 4
    $\begingroup$ Hint: $$b_n\xrightarrow[n\to\infty]{}b\implies \frac{b_1+\ldots+b_n}n\xrightarrow[n\to\infty]{}b$$ $\endgroup$ – DonAntonio Jun 6 '13 at 23:30
  • $\begingroup$ This is called Cesaro summation. $\endgroup$ – Julien Jun 7 '13 at 0:21
  • $\begingroup$ gosh dang it, it is a cesaro sum!!! Well, I know which proof to mimick now, thank you. $\endgroup$ – DaveNine Jun 7 '13 at 0:44

For large enough $j$, $S_j$ is within $\epsilon$ of $\frac{3\pi}{4}$. Make this rigorous by quantifying what "large enough" means. Then $T_n$ is an average of a "handful" of terms that are not necessarily close to $\frac{3\pi}{4}$ and an unbounded finite number of terms that are within $\epsilon$ of $\frac{3\pi}{4}$. Again, make this rigorous (especially for an analysis qual!), use inequalities, group like terms together and you should have something like $$\left|T_n-\frac{\overbrace{\left(\frac{3\pi}{4}+\cdots+\frac{3\pi}{4}\right)}^{n\text{ copies}}}{n}\right|\leq\frac{\text{bounded sum}+(n-n_{\epsilon})\epsilon}{n}\to\epsilon$$


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