Series about inserting parentheses

Question

Problem : Suppose that the series $\sum_{}^{} a_k$ converges and ${n_j}$ is a strictly increasing sequence of positive integers. Define the sequence $b_k$ as follows :

$b_1=a_1+...+a_{n_{1}}$

$b_2=a_{n_1+1}+...+a_{n_{2}}$

...

$b_k=a_{n_{k-1}+1}+...+a_{n_{k}}$

Prove that $\sum_{}^{} b_k$ converges and that $\sum_{}^{} a_k=$ $\sum_{}^{} b_k$

My Proof :

Let $s_m=\sum_{i=1}^{n_m} a_i$ $\forall m \in \mathbb{N}$ then by definition, $s_m=\sum_{i=1}^{n_m} a_i=\sum_{k=1}^{m} b_k$

Since $n_j$ is strictly increasing sequence of positive integers, $\lim_{m\to\infty}\ n_m= \infty$

Since $\lim_{m\to\infty}\ n_m= \infty$ and $\sum_{}^{} a_k$ converges, $\lim_{m\to\infty}\ s_m$ exists and

$\lim_{m\to\infty}\ s_m=$ $\lim_{m\to\infty}\sum_{i=1}^{n_m} a_i=$ $\sum_{}^{} a_k$

Since $\forall m \in \mathbb{N}$ $s_m=\sum_{k=1}^{m} b_k$ and $\lim_{m\to\infty}\ s_m$ exists,

$\sum_{}^{} b_k$ converges and $\sum_{}^{} a_k$=$\sum_{}^{} b_k$

But, I'm not sure my proof is correct. Please give me feedback on my proof.

Answer 1

Yes, the idea of your proof is fine. However, your argument in the 2nd and 3rd line is unclear (among other things since $n_j$ is not defined). I suppose, in short your argument is:

Define $s_n := \sum_{k = 0}^n a_k$ and $s_m := \sum_{k = 0}^m b_k$. Then $(s_m)_{m \in \mathbb{N}}$ is a subsequence of $(s_n)_{n \in \mathbb{N}}$. Since $\sum a_k$ converges (by definition) $(s_n)_{n \in \mathbb{N}}$ converges, so does $(s_m)_{m \in \mathbb{N}}$ and thus $\sum b_k$ converges. Since limits are unique, they coincide.

Make those two lines precise and you are good to go.