Convergent Series and Partial Sums

[H12504106]

Homework Statement

Let \sum_{n=1} a_n and \sum_{n=1} b_n be convergent series. For each n \in \mathbb{N}, let c_{2n-1} = a_n and c_{2n} = b_n. Prove that \sum_{n=1} c_n converges.

Homework Equations

The Attempt at a Solution

Not sure whether the following solution is correct or not.
Let S_n, T_n, R_n be the partial sums of the series \sum_{n=1} a_n, \sum_{n=1} b_n, \sum_{n=1} c_n respectively. Now (R_{2n-1}) = c_1 + c_2 +...+ c_{2n-1} = (a_1 +...+ a_n)+ (b_1 +...+b_{n-1}) = S_n +T_{n-1}. Similarily, (R_{2n}) = c_1 + c_2 +...+ c_{2n-1} + c_{2n} = (a_1 +...+ a_n)+ (b_1 +...+b_n) = S_n +T_n. Since \sum_{n=1} a_n[/itex] and \sum_{n=1} b_n converges, the sequence (S_n) and (T_n) converges. Since (R_{2n-1}) and (R_{2n}) converges to the same value, (R_n) converges. Hence, the series \sum_{n=1} c_n converges.

 

[LCKurtz]

Homework Statement

Let \sum_{n=1} a_n and \sum_{n=1} b_n be convergent series. For each n \in \mathbb{N}, let c_{2n-1} = a_n and c_{2n} = b_n. Prove that \sum_{n=1} c_n converges.

The Attempt at a Solution

Not sure whether the following solution is correct or not.
Let S_n, T_n, R_n be the partial sums of the series \sum_{n=1} a_n, \sum_{n=1} b_n, \sum_{n=1} c_n respectively. Now (R_{2n-1}) = c_1 + c_2 +...+ c_{2n-1} = (a_1 +...+ a_n)+ (b_1 +...+b_{n-1}) = S_n +T_{n-1}. Similarily, (R_{2n}) = c_1 + c_2 +...+ c_{2n-1} + c_{2n} = (a_1 +...+ a_n)+ (b_1 +...+b_n) = S_n +T_n. Since \sum_{n=1} a_n[/itex] and \sum_{n=1} b_n converges, the sequence (S_n) and (T_n) converges. Since (R_{2n-1}) and (R_{2n}) converges to the same value, (R_n) converges. Hence, the series \sum_{n=1} c_n converges.

<br /> <br /> The assertions you make look to be all true. But I think you need to give a more complete explanation for the last two sentences, because proving it carefully is essentially the same as the original problem. I would think along the lines if Σ a_n = S and Σb_n = T, you should be able to show directly that the c series converges to S + T with an ε, N argument.