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A more clear statement:

Suppose $a_k \to 0$. The series $\sum_{k=1}^{\infty} a_k$ converges iff $\sum_{k=1}^{\infty} (a_{2k}+a_{2k+1})$ converges.

I have been able to prove the forward direction. I'm stuck on backward direction.

3 Answers 3

0

Let $\varepsilon>0$. Then there exists $n_0$ such that, for all $m>n$, $ \left|\sum_{k=n_0}^ma_{2k}+a_{2k+1}\right|<\varepsilon. $ There also exists $n_1$ such that $|a_k|<\varepsilon$ if $k\geq n_1$. Let $n=\max\{n_0,n_1\}$.

Then, for any $m>2n$ and a proper $c\in\{0,1\}$, $ \left|\sum_{k=2n}^ma_k\right|=\left|\left(\sum_{k=n}^{\lfloor m/2\rfloor} a_{2k}+a_{2k+1}\right) - c\,a_{2\lfloor m/2\rfloor+1}\right|\leq\left|\sum_{k=n}^{\lfloor m/2\rfloor} a_{2k}+a_{2k+1}\right| +\left| \,a_{2\lfloor m/2\rfloor+1}\right|<2\varepsilon. $ So the tails of the series $\sum a_k$ go to zero, and so the series is convergent.

1

Hint: Because $\sum (a_{2k}+a_{2k+1})$ converges, the sequence of partial sums of this series is Cauchy.

Conclude from this that the sequence of partial sums of the series $\sum a_n$ is Cauchy. For odd $m$ and/or even $n$, where $m\lt n$, we will need the fact that for large $m$ and $n$, the terms $a_m$ and $a_n$ have absolute value close to $0$. This part follows from the assumption that the $a_n$ have limit $0$.

0

Suppose that $\sum_{k\ge 1}(a_{2k}+a_{2k+1})=L$; clearly we want to show that $\sum_{k\ge 1}a_k=L+a_1$. Let $\epsilon>0$; there is an $m_\epsilon\in\Bbb N$ such that $\left|L-\sum_{k=1}^n(a_{2k}+a_{2k+1})\right|<\epsilon$ whenever $n\ge m_\epsilon$. There is also an $m_\epsilon'\in\Bbb N$ such that $|a_k|<\epsilon$ whenever $n\ge m_\epsilon'$.

Now consider a partial sum $\sum_{k=1}^na_k$. Suppose first that $n$ is odd; say $n=2m+1$. Then $\sum_{k=1}^na_k=a_1+\sum_{k=1}^m(a_{2k}+a_{2k+1})\;,$ so if $m\ge m_\epsilon$ we have

$\left|a_1+L-\sum_{k=1}^na_k\right|=\left|L-\sum_{k=1}^m(a_{2k}+a_{2k+1})\right|<\epsilon\;.$

Now suppose that $n$ is even; say $n=2m$. Then $\sum_{k=1}^na_k=a_1+\sum_{k=1}^{m-1}(a_{2k}+a_{2k+1})+a_{2m}\;,$ so if $m-1\ge m_\epsilon$ and $n\ge m_\epsilon'$ we have

$\begin{align*} \left|a_1+L-\sum_{k=1}^na_k\right|&=\left|L-\sum_{k=1}^{m-1}(a_{2k}+a_{2k+1})-a_n\right|\\ &\le\left|L-\sum_{k=1}^{m-1}(a_{2k}+a_{2k+1})\right|+|a_n|\\ &<\epsilon+\epsilon\\ &=2\epsilon\;. \end{align*}$

In all cases, therefore, if $n\ge\max\{m_\epsilon,m_\epsilon'\}$, then

$\left|a_1+L-\sum_{k=1}^na_k\right|<2\epsilon\;,$

and it follows that $\sum_{k\ge 1}a_k=a_1+L\;.$