My intuition says yes, but I'm trying to prove it using limit laws and getting stuck. So far I have the following:

Let $(s_n)$ be the sequence of partial sums of $\sum(a_n+b_n)$ and $(t_n)$ be the sequence of partial sums of $\sum(a_n-b_n)$. So $\lim s_n=\sum(a_n+b_n)$ and $\lim s_n=\sum(a_n-b_n)$. Then $\lim(s_n+t_n)=\lim s_n+\lim t_n$ (because $(s_n),(t_n)$ converge), and then $\lim(s_n+t_n)=\lim((a_n+b_n)+(a_n-b_n))=\lim(2a_n)=\sum(a_n+b_n)+\sum(a_n-b_n)$.

But I'm not convinced that actually goes anywhere, or if I'm taking the right approach. If anyone has any ideas to help in making this proof work out I would really appreciate it.


1 Answer 1


Hint: Note that $$a_n=\frac{(a_n+b_n)+(a_n-b_n)}{2}$$ and $$b_n=\frac{(a_n+b_n)-(a_n-b_n)}{2}.$$

  • $\begingroup$ I'm so sorry, but I have no idea what to do with that 😭😭 where is that useful in the proof? $\endgroup$
    – ksea
    Oct 14, 2019 at 1:18
  • 2
    $\begingroup$ This allows you to write $\sum a_n$ and $\sum b_n$ as linear combinations of things that you know converge, $\sum a_n+b_n$ and $\sum a_n-b_n$. And the limit laws are all about how limits 'just work' with linear combinations. $\endgroup$ Oct 14, 2019 at 1:53
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    $\begingroup$ @ksea Let $A_n$ denote the partial sum of $\sum a_n$. Then $A_n=\frac12(s_n+t_n)$. Since $s_n$ and $t_n$ are convergent, $A_n$ is convergent, so $\sum a_n$ is convergent. Similarly we can get the convergence of $\sum b_n$. $\endgroup$
    – Feng
    Oct 14, 2019 at 1:59

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