# Convergent Series and Partial Sums

by H12504106
Tags: convergent, partial, series, subsequence, sum
 P: 6 1. The problem statement, all variables and given/known data 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. 2. Relevant equations 3. 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 [TEX]\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.
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P: 6,759
 Quote by H12504106 1. The problem statement, all variables and given/known data 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. 3. 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 [TEX]\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.
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 an = S and Σbn = T, you should be able to show directly that the c series converges to S + T with an ε, N argument.

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