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Exercise 38, Chatper 8 of Apostol's Mathematical Analysis. Given a monotonic sequence $\{a_n\}$ of positive terms, such that $\lim_{n\to\infty}a_n=0$. Let $$s_n=\sum_{k=1}^na_k,\quad u_n-\sum_{k=1}^n(-1)^ka_k,\quad v_n=\sum_{k=1}^n(-1)^ks_k.$$

Prove that:

a) $v_n=\frac12u_n+(-1)^ns_n/2$.

I've shown this without any difficulty.

b) $\sum_{n=1}^\infty(-1)^ns_n$ is $(C,1)$ summable and has Cesaro sum $\frac12\sum_{n=1}^\infty (-1)^na_n$.

My attempt so far: By the conditions on $\{a_n\}$, $\sum_{n=1}^\infty(-1)^na_n$ converges by the alternating series test. Let $s=\sum_{n=1}^\infty(-1)^ka_k$. Note $\lim_{n\to\infty} \frac1n\sum_{k=1}^nu_k=s$, by Cesaro's theorem. Let $\sigma_n=\frac1n\sum_{k=1}^nv_k$. By part a)

$$\sigma_n=\frac1n\sum_{k=1}^nv_k=\frac1{2n}\left(\sum_{k=1}^nu_k+\sum_{k=1}^n(-1)^ks_k\right)=\frac1{2n}\sum_{k=1}^nu_k+\frac1{2n}v_n.$$

So, $$\lim_{n\to\infty}\frac1n\left(\sum_{k=1}^{n-1}v_k+\frac12v_n\right)=\lim_{n\to\infty}\frac1{2n}\sum_{k=1}^nu_k=\frac12s.$$

So, I need to show that $$\lim_{n\to\infty}\frac1n\left(\sum_{k=1}^{n-1}v_k+\frac12v_n\right)=\lim_{n\to\infty}\frac1n\sum_{k=1}^\infty v_k$$.

This seems reasonable enough, but I'm not finding a way to do it.

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Since $a_n \to 0$, so does $A_n:= \frac{s_n}{n} \to 0$. Since

$$v_{2n} = s_2 - s_1 + s_4 - s_3 + \cdots + s_{2n} - s_{2n-1} = a_2 + a_4 + \cdots + a_{2n},$$

we have

$$\frac{v_{2n}}{2n} < A_{2n} \to 0.$$

Also,

$$v_{2n+1} = v_{2n} - s_{2n+1} = -a_1 - a_3 - \cdots - a_{2n+1}.$$

Thus

$$\left|\frac{v_{2n+1}}{2n+1}\right| < A_{2n+1} \to 0.$$

Consequently,

$$\lim_{n\to \infty} \frac{v_n}{n} = 0.$$

Hence,

$$\lim_{n\to \infty} \frac{1}{n}\sum_{k = 1}^{n-1} v_k = \frac{1}{2}s,$$

which implies

$$\lim_{n\to \infty} \sigma_n = \lim_{n\to \infty} \frac{1}{n+1}\sum_{k = 1}^n v_k \cdot \frac{n+1}{n} = \frac{1}{2}s\cdot \lim_{n\to \infty} \frac{n+1}{n} = \frac{1}{2}s.$$

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  • $\begingroup$ Thanks, kobe. Nice answer. $\endgroup$ – Tim Raczkowski Feb 27 '15 at 18:41

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