# Convergence of series

1. May 5, 2013

### drawar

1. The problem statement, all variables and given/known data

Let $(a_n)$ be a sequence.
(i) Prove that if $\sum\limits_{n = 1}^\infty {{a_n}}$ converges, then $\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)}$ also converges.

(ii) Prove that if $\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)}$ converges and $a_n \to 0$, then $\sum\limits_{n = 1}^\infty {{a_n}}$ converges.

2. Relevant equations

3. The attempt at a solution

(i) Let ${R_n} = \sum\limits_{k = 1}^n {{a_k}}$, ${S_n} = \sum\limits_{k = 1}^n {{a_{2k - 1}}}$, and ${T_n} = \sum\limits_{k = 1}^n {{a_{2k}}}$

Then ${R_{2n}}={S_n}+{T_n}$, since $\sum\limits_{n = 1}^\infty {{a_n}}$ converges, the sequence $(R_n)$ converges, and so is the subsequence $(R_{2n})$. It follows that $\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)}$.

(ii) Ok so I'm stuck on this part. I already have $(R_{2n})$ converges and $(a_n)$ is bounded, how can I go about proving that $(R_{n})$ converges as well? Thank you!

2. May 5, 2013

### Staff: Mentor

The convergence of Rn does not guarantee the convergence of Sn or Tn, you cannot write it as sum like that.
The idea to use a sequence and subsequences is good, however.

(ii) how can R2n+1 deviate from R2n, if an->0?

3. May 8, 2013

### drawar

I'm sorry I wasn't able to reply earlier.

For (i), I agree with you that the convergence of $R_n$ does not imply the convergence of $S_n$ or $T_n$ but it can imply that ($S_n$+$S_n$) converges right?

Btw, I think I may get your point for part (ii), let me just say what I'm thinking: Since $R_{2n}$ = $R_{2n-1}$+$a_{2n}$, it can be deduced that $R_{2n-1}$ converges to the same limit as $R_{2n}$, as a result, $R_{n}$ converges.

4. May 8, 2013

### Staff: Mentor

If you replace the second S by a T (typo?), yes.

That was the idea I had in mind, indeed.