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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?

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  • $\begingroup$ Just look at $\dfrac{c_n}{n}$ and see if it tends to zero? $\endgroup$
    – dezdichado
    Mar 16, 2018 at 5:12
  • $\begingroup$ Yeah, I definitely understand this, but I just didn't get what the book meant there @dezdichado $\endgroup$ Mar 16, 2018 at 5:14
  • $\begingroup$ you are mistaken. $\sigma_n - \dfrac{n-1}{n}\sigma_{n-1}$ is indeed equal to $\dfrac{c_n}{n}.$ Your challenge would be to prove that the left hand side tends to $0$, provided that $\sigma_n\to \sigma,$ some finite number. $\endgroup$
    – dezdichado
    Mar 16, 2018 at 5:17
  • $\begingroup$ @dezdichado This is how I do that: $\sigma_n=\frac {s_1+\cdots +s_n}{n}, \sigma_{n-1}=\frac {s_1+\cdots +s_{n-1}}{n-1} $ so, $\frac {s_1+\cdots +s_n}{n}-\frac{n-1}{n}\frac {s_1+\cdots +s_{n-1}}{n-1}=\frac {s_n}{n}$ How do you get $\frac {c_n}{n}$? $\endgroup$ Mar 16, 2018 at 5:24
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    $\begingroup$ never mind, I got it wrong. $\endgroup$
    – dezdichado
    Mar 16, 2018 at 6:28

2 Answers 2

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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 $\sum_{k=1}^{\infty} c_k$.

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  • $\begingroup$ By the way, there is another question about this topic: Again we need to prove that " if $\sum c_n$ is Cesaro summable to $\sigma$, then $\sum c_n$ converges to $\sigma$ "And the author proves that $s_n-\sigma_n\rightarrow 0$ and concludes the proof with that. Showing $s_n-\sigma_n\rightarrow 0$ instead of $\sigma_n\rightarrow \sigma$ is somehow serving the same purpose? $\endgroup$ Mar 16, 2018 at 6:38
  • $\begingroup$ I think you are referring to Hardy's Tauberian Theorem. You need an extra condition on $c_n$ to show $s_n-\sigma_n \to 0$. You can get proofs of Hardy's theorem on the net easily. $\endgroup$ Mar 16, 2018 at 8:36
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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 $\sigma_7=\frac {1}{7}(6\sigma_6+s_7)=\frac {1}{7}(0+(-4/2))=\frac {-4}{14}.$

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