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Since I'm studying real analysis using this book by myself, I'm not sure whether or not my method to prove convergence of sequence is right. I'm working on the above question's (d), and my solution was kind of different from the solution provided by the following website. The solution in the website is much more beautiful and more concise than mine, but I believe my method also works.
http://minds.wisconsin.edu/handle/1793/67009

3.14 If $\{s_n\}$ is a complex sequence, define its arithmetic means $\sigma_n$ by

$$\sigma_n = \dfrac{s_0+s_1+...+s_n}{n+1} (n=0,1,2,...).$$

(d) Put $a_n=s_n-s_{n-1}$, for $n\geq 1$. Show that

$$s_n-\sigma_n=\dfrac{1}{n+1}\sum_{k=1}^n k a_n$...$(1)$$

Assume that lim$(n a_n)=0$ and that ${\sigma_n}$ converges. Prove that $\{s_n\}$ converges.

My solution:

Since $\sigma_n$ converges, it is a Cauchy sequence. So,

$\forall\epsilon,\forall\delta>0, \exists N\in\mathbb{N}$ s.t.$\forall n, \forall m> N , |n a_n-0|<\epsilon$ and $|\sigma_n-\sigma_m|<\delta$

Let $M=\max\{k |a_k|:1\leq k\leq m\}$

From eq. $(1)$,

\begin{align} |s_n-s_m|&=|(\sigma_n-\sigma_m)+(\dfrac{1}{n+1}\sum_{k=1}^{n}k a_k-\dfrac{1}{m+1}\sum_{k=1}^{m}k a_n)|\\&\leq|\sigma_n-\sigma_m|+\dfrac{|(m+1)(a_1+2a_2+...+n a_n)-(n+1)(a_1+2a_2+...m a_m)|}{(n+1)(m+1)}\\&=|\sigma_n-\sigma_m|+\dfrac{|(m+1)((m+1)a_{m+1}+(m+2)a_{m+2}+...+n a_n)+(m-n)(a_1+2a_2+...m a_m)|}{(n+1)(m+1)}\\&<\delta+\dfrac{(m+1)(n-m)\epsilon+(m-n)mM}{(n+1)(m+1)}\\&=\delta+\dfrac{(n-m)\epsilon}{n+1}+\dfrac{(m-n)mM}{(n+1)(m+1)}\end{align}

As $\epsilon, \delta \to 0, n \to \infty$ and $m \to n$, the right side converges to $0$; therefore, the Cauchy sequence $\{s_n\}$ converges.

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Am I allowed to assume that M such that $M=max\{k|a_k|:1≤k≤m\}$ exists? –  Aran Komatsuzaki May 20 at 0:53

1 Answer 1

up vote 3 down vote accepted

You're overcomplicating things. Suppose that $\sigma_n\to s$. I claim $s_n\to s$.

Cesaro's theorem says that if $x_n\to x$, then $\widehat{x_n}\to x$ where $\widehat{x_n}=\displaystyle \frac{1}{n+1}\sum_{k=0}^nx_k$.

Let $x_k=ka_k$. Then $ka_k\to 0$; so $\widehat{x_k}=s_k-\sigma_k\to 0$. Since $\sigma_k\to s$, $s_n\to s$.

Thus, all you have to do is prove Cesaro's theorem. You can assume that $x_n\to 0$ (why?), and then it is all pretty easy.

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Thanks for your suggestion. Although I didn't know this theorem before, surprisingly, I unconsciously proved a part of the theorem in the question (a). I'm really glad to know a simpler way to solve the question. –  Aran Komatsuzaki May 20 at 1:07

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