# suppose $\sum a_n$ converges. Is it true that then $\sum (-1)^na_n$ will

suppose $\sum a_n$ converges. Is it true that then $\sum (-1)^na_n$ will also converge.

I think that the statement is true but I'm having trouble proving it.

• Do you know about absolute convergence? – Randall Apr 5 '18 at 19:44
• so since it converges absolutely then $\sum (-1)^na_n$ will also converge? @Randall – Skrrrrrtttt Apr 5 '18 at 19:47
• abs cov says if $\sum_n |a_n|$ converges then so does $\sum_n a_n$. That's exactly your situation. – Randall Apr 5 '18 at 19:48
• @Randall that's not the situation because $a_{n}$ converging does not imply $|a_{n}|$ converging. – pwerth Apr 5 '18 at 19:50
• @Randall $|a_n| \neq a_n$. – Mesmerized student Apr 5 '18 at 19:52

## 4 Answers

Let consider

• $a_n=(-1)^{n}\frac1n\implies \sum a_n$ converges

but

• $\sum (-1)^na_n=\sum \frac1n$ which diverges

No, this is in general wrong. Consider for instance $a_n = \frac{(-1)^n}{n}$.

Harmonic series can be counterexample for your statement. So no, it’s not true.

$a_k= \frac 1k$ if $k$ even $a_k= -\frac 1{k-1}$ is $k$ odd

$\sum (-1)^ka_k = \sum \frac 1{k}$ diverges but $\sum a_k = \frac1{2n}$ converges

Your result is true is all $a_k$ are positive in the first place or if the |a_k| serie converges