Absolute and conditional convergence of series proof

Question

Assume $ \sum_{n=1}^{\infty}a_{n} $ is absolutely convergent.

and assume $ \sum_{n=1}^{\infty}b_{n}$ is conditionally convergent.

What can we say for sure about $ \sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right) $ ?

I'm thinking that for sure, we could say that the series is convergent. But the series will not converge absolutely.

To prove the convergence is not hard I guess, we just use limit arithmetic over the sum sequences aka $ S_{p}=\sum_{n=1}^{p}b_{n},T_{p}=\sum_{n=1}^{p}a_{p} $.

But I'm not sure how to show that the series will not converge absolutely (is it true at all)? because $ |a_{n}+b_{n}|\leq|a_{n}|+|b_{n}| $ we know that $ \sum_{n=1}^{\infty}|b_{n}| $ diverges, but we cannot really say anything based on comparison test.

I'd appreciate some help. Thanks in advance

Answer 1

Use proof by contradiction. Suppose $\sum (a_n+b_n)$ converges absolutely. Then so does $\sum b_n$ because $|b_n| \leq |a_n+b_n|+|a_n|$ which leads to a contradiction.

Answer 2

You can also argue using the Riemann rearrangement theorem.

Suppose $\sum_n a_n = A$ and $\sum_n b_n = B$.

Since $\sum_n b_n$ is conditionally convergent, we can find a rearrangement $\sum_k b_{n(k)}$ that converges to some other limit $B' \neq B$. (Here $k \mapsto n(k)$ is a bijection from $\mathbb N$ to itself.)

Then $\sum_n (a_n + b_n) = A + B$, whereas $\sum_k (a_{n(k)} + b_{n(k)}) = A + B' \neq A+B$. (Note that $\sum_n a_n = \sum_k a_{n(k)}$ because $\sum a_n$ is absolutely convergent, hence any rearrangement converges to the same limit $A$.)

This shows that we can change the value of $\sum_n(a_n+b_n)$ by rearranging the terms, hence the series is not absolutely convergent.