# If both $\sum(a_n+b_n)$ and $\sum (a_n-b_n)$ converge, do $\sum a_n$, $\sum b_n$ converge?

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.

Hint: Note that $$a_n=\frac{(a_n+b_n)+(a_n-b_n)}{2}$$ and $$b_n=\frac{(a_n+b_n)-(a_n-b_n)}{2}.$$
• 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. Commented Oct 14, 2019 at 1:53
• @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 $\sum b_n$.