$\sum_{n=1}^{\infty}(a_{2n-1}+a_{2n})$ is convergent and $\lim_{n\to\infty}a_n=0$. What about $\sum_{n=1}^{\infty}a_n$? 
Suppose that $$\sum_{n=1}^{\infty}(a_{2n-1}+a_{2n})$$ is convergent, and $$\lim_{n\to\infty}a_n=0.$$ Prove that $\sum_{n=1}^{\infty}a_n$ is convergent.

I know if a series is convergent then $\lim_na_n=0$, but I cannot say that $\lim_na_n=0$ implies convergence of the series . It seems nothing helps for this problem. 
How to solve this question?  
 A: I really don't understand the other answer... Anyway,
Define $\displaystyle S_n=\sum_{k=1}^na_k$.
Our goal is to prove that the sequence $(S_n)$ converges.


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*We have that $\displaystyle S_{2n}=\sum_{k=1}^{2n}a_k=\sum_{k=1}^n(a_{2k-1}+a_{2k})$


Therefore the sequence $(S_{2n})$ converges.


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*It remains to prove that $(S_{2n+1})$ converges and $\displaystyle \lim_{n\to \infty}S_{2n+1}=\lim_{n\to \infty}S_{2n}$


See that $\displaystyle S_{2n+1}=S_{2n}+a_{2n+1}$
Since $\lim_{n\to \infty}a_{2n+1}=0$, we deduce the existence of the limit and  $\displaystyle \lim_{n\to \infty}S_{2n+1}=\lim_{n\to \infty}S_{2n}$


*

*This proves the convergence of the sequence $(S_n)$ which turn implies the convergence of $\sum_0^\infty a_k$

A: $$N=2M,\ n>N:\ \sum_{n=1}^{\infty}a_n=\lim_{n\to\infty}(S_n+a_{n+1})=\sum_{n=1}^{\infty}(a_{2n-1}+a_{2n}+0)=S_n$$
Note that the third term is because $\lim_{n\to\infty}a_n=0$. 
Or you can compare this counter example: $a_n=(-1)^n$, it does not satisfy $\lim_{n\to\infty}a_n=0$, hence although $\sum_{n=1}^{\infty}(a_{2n-1}+a_{2n})$ is convergent $\sum_{n=1}^{\infty}a_n$ is still divergent.
