Cesàro sum of the alternating harmonic series

Question

Consider $$ \sum^{\infty}_{k=1}(-1)^{k-1}\frac{1}{k} \tag{a} $$ where the Cesàro sum is $\lim_{k\to\infty}\frac{\sum^k_{j}s_j}{k}$ where $s_k$ is the $k$th partial sum. Is $(a)$ Cesàaro convergent?

We know that the alternating harmonic series converges (by the "normal") definition to $\ln2$, I am therefore tempted to say that the Cesàro sum is therefore 0 because $\lim_{k\to\infty}\ln2/k=0$. We know that if a series that consists on non-negative terms converges normally, then it is also Cesàro convergent, unfortunately that is not the case here. Any ideas?

Answer 1

If a series is convergent and its sum is $A$, then its Cesàro sum is convergent with same value $A$. So the limit is $$ \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n s_k = \ln 2 $$ where $s_k = \sum_{\ell=1}^k \frac{(-1)^{\ell-1}}{\ell}$ is the partial sum of the original series. (You seem to have the definition of Cesàro summation a bit confused in your question, the fact that $(\ln 2)/k\to 0$ is not relevant.)

Answer 2

Any sequence which converges also converges in Cesaro sense. Since $(s_n)$ converges so does $\frac {\sum _{k=1}^{n} s_k} n$ and the limit is same as the limit of $(s_k) $ namely $\ln 2$.

Answer 3

If a series converges, the Cesaro sum, which is nothing but the arithmetic mean of the partial sums, will converge to the same value. This is because the more terms you add, the more these terms get close to the limit sum, and so does the average.

To illustrate, consider a geometric sum of common ratio $\dfrac1{10}$, giving the partial sums

$$1,1.1,1.11,1.111,1.1111,1.11111,\cdots$$

The partial Cesaro sums are

$$1,1.05,1.07,1.08025,1.08642,1.095679014,\cdots$$

and it is clear that more and more terms will be closer and closer to $\dfrac{10}9$, making the average tend to that value.