# If $\sum_{n=1}^\infty a_n$ converges to $A$, then ${1 \over 2}(a_1 + a_2) + {1 \over 2}(a_2 + a_3) + \cdots$ converges

I need to prove

If $a_1 + a_2 + a_3 + \cdots$ converges to $A$, then ${1\over2}(a_1 +a_2) + {1\over2}(a_2 + a_3) + {1\over2}(a_3+a_4) + \cdots$ converges. Find what the latter series converges to.

PROOF: Given that $\displaystyle \sum_{n=1}^\infty a_n$ converges to $A$, then the sequence of partial sums of the series, $\{s_n\}$, converges to $A$, where $\displaystyle s_n = \sum_{k=1}^na_k$. That is, $$\lim_{n\to\infty}s_n = A.$$ Now, for the $2^{nd}$ series, observe that we may describe its sequence of partial sums as $\{t_n\}$, where $\displaystyle t_n = {1\over2}\sum_{k=1}^n\left(a_k + a_{k+1}\right)$. Then \begin{align}\lim_{n\to\infty}t_n &= \lim_{n\to\infty}\left({1\over2}\sum_{k=1}^n(a_k + a_{k+1})\right) \\ &= {1\over2}\lim_{n\to\infty}\left(\sum_{k=1}^na_k + \sum_{k=1}^na_{k+1}\right) \tag{*}\\ &= {1\over2}\lim_{n\to\infty}\sum_{k=1}^na_k + {1\over2}\lim_{n\to\infty}\sum_{k=1}^na_{k+1}\\&= {1\over2}\lim_{n\to\infty}s_n + {1\over2}\lim_{n\to\infty}s_{n+1} \tag{\bf{\star}}\\ &= {1\over2}A + {1\over2}A \\&= A.\end{align} Thus, the sequence of partial sums, $\{t_n\}$, converges and hence the series ${1\over2}(a_1 +a_2) + {1\over2}(a_2 + a_3) + {1\over2}(a_3+a_4) + \cdots$ converges to $A$. $\square$

QUESTION: I may split the sequence of partial sums at $(*)$ up because it is a finite sum, correct? If it were an infinite series, then I don't think it would be valid since we don't necessarily know that it converges to begin with (which is what we're trying to show). Also, is it correct to write $\lim_{n\to\infty}s_{n+1}$ at $(\star)$? Or is it supposed to be $\lim_{n\to\infty}s_n$?

• The sequence $t_n$ should converge to $A-\frac{1}{2}a_1$. – edm Oct 18 '16 at 10:13
• @edm, I see why this is now, so thanks for catching that. It is because: \begin{align}\sum_{k=1}^na_{k+1} &= (-a_1 + a_1) + a_2+a_3 + \cdots + a_n + a_{n+1} \\ &= -a_1 + s_{n+1} \\&= s_{n+1} - a_1\end{align} – Decaf-Math Oct 18 '16 at 10:28
• Wouldn’t $$s_{n+1} = \sum_{k=1}^{n+1}a_k,$$ instead of $$\sum_{k=1}^n a_{k+1}.\tag*{?}$$ The latter would simply equal $s_{n+1} - a_1$ if that’s the case. This supports what @edm said. – Mr Pie Jan 29 '18 at 3:29

You are allowed to rearrange the terms of an infinite sum only if it's absolutely convergent. Do you have some information regarding $a_n$? Do you know, for example, that $a_n\ge0$ definitely?
For the second question: it is true that $$\lim_{n\to+\infty}s_n=\lim_{n\to+\infty}s_{n+1}.$$
If $\sum_{n=1}^\infty a_n$ converges to $A$,
then ${1 \over 2}(a_1 + a_2) + {1 \over 2}(a_2 + a_3) + \cdots$ can be rewritten as :
${1 \over 2}(a_1) + {1 \over 2}(a_2 + a_2) + {1 \over 2}(a_3 + a_3) + \cdots ={1 \over 2}(a_1) + a_2 + a_3 + \cdots =$A$-{1 \over 2}a_1$
if $a_n$ is a sequence of positive terms