Proving Convergence of Series: (a_n) and (a_{2n-1} + a_{2n})

[drawar]

Homework Statement

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.

Homework Equations

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!

 

[mfb]

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

(ii) how can R_2n+1 deviate from R_2n, if a_n->0?

 

[drawar]

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

(ii) how can R_2n+1 deviate from R_2n, if a_n->0?

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.

 

[mfb]

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?

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

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.

That was the idea I had in mind, indeed.