Take away a third at a time

Calculus Level 3

$\large 1 + 2 - 3 + 4 + 5 - 6 + 7 + 8 - 9 + \cdots$

The series $\displaystyle \sum_{j=1}^{\infty} a_j$ is said to be Cesàro summable , with Cesaro Sum $A$ , if the average value of its partial sums $\displaystyle s_k=\sum_{j=1}^k a_j$ tends to $A$ , meaning that $\displaystyle A=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^ns_k$ .

Is the series above Cesàro summable?

No Yes

1 solution

Otto Bretscher
May 15, 2016

This is a clever problem, Comrade! My first intuition proved to be wrong here ;)

We can show by induction that we have the partial sums $s_{3n}=3\sum_{k=1}^{n-1}k=\frac{3(n-1)n}{2}$ , so that $A_{3n}>\frac{s_{3n}}{3n}=\frac{n-1}{2}$ for the average value $A_n=\frac{1}{n}\sum_{k=1}^{n}s_k$ of the partial sums. Since $A_{3n}$ diverges, the series fails to be Cesàro summable. The answer is $\boxed{No}$

Take away a third at a time

1 solution

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