Show that $\sum c_n$ is not Cesaro summable

Question

Consider the sequence $c_n=(-1)^{n-1}n$
Show that $\sum c_n$ is not Cesaro summable using the hint: "If $\sum c_n$ is Cesaro summable, then $\frac {c_n}{n}$ tends to $0$"

the $N^{th}$ Cesaro sum of the series $\sum_{k=1}^{\infty}c_k$ is $\sigma_N=\frac {s_1+\cdots s_N}{N}$, where $s_n=\sum_{k=1}^{n}c_k$

So $\sigma_n-\frac{n-1}{n}\sigma_{n-1}= \frac {s_n}{n}$ but in the solution manual it says that $\sigma_n-\frac{n-1}{n}\sigma_{n-1}= \frac {c_n}{n} \tag {*}$ and concludes that $\lim_{n\rightarrow \infty} \frac {(-1)^{n-1}n}{n}\neq 0$ , which I agree..
But I didn't understand the starred part.
By the way are the notations $\sum c_n$ and $\sum_{k=1}^{\infty}c_k$ the same?

Answer 1

I believe there is a misprint and it is $\frac {s_n} n$ that must tend to $0$. Now look at $s_n$, say for odd n and group the terms 2-by-2. You can easily see that $\frac {s_n} n$ does not tend to $0$ so $\sum c_n$ is not Cesaro summable. And yes, $\sum c_n$ is just a short form for $\sum_{k=1}^{\infty} c_k$.

Answer 2

By direct calculation we have

$s_{2N}=\sum_{n=1}^{2N}c_n=N/2.$

$s_{2N-1}=\sum_{n=1}^{2N-1}=-N/2.$

$\frac {1}{2m}\sigma_{2m}=0.$

$\frac {1}{2m-1}\sigma_{2m-1}=\frac {-m}{4m-2}.$

For example $s_6=(1-2)+(3-4)+(5-6)=-3.$ And $s_7=s_6+c_7=(-3)+7=4.$

And $\sigma_6=$ $=\frac {1}{6}((s_1+s_2)+(s_3+s_4)+(s_5+s_6))=\frac {1}{6}((1/2-1/2)+(2/2-2/2)+(3/2-3/2))=0.$ And $\sigma_7=\frac {1}{7}(6\sigma_6+s_7)=\frac {1}{7}(0+(-4/2))=\frac {-4}{14}.$