1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.