# If $\sum a_n$ is convergent and $a_n>0$, then $\sum (-1)^n a_n$ is convergent.

If $\sum a_n$ is convergent and $a_n>0$, then $\sum (-1)^n a_n$ is convergent.

So far I've tried convergence/divergence tests and also I tried to prove this using partial sums. But tests do not work because the latter series includes negative terms. Maybe the Alternating Series Test could have worked but "$a_n$ is decreasing" is not an initial condition. I know the proof will be very easy but I am stuck here. Could you provide me with a little hint?

• The hint of Nicolò Ruggeri is the good one. Commented Feb 23, 2016 at 14:02
• The sum of odd and even terms (separately) converge, so you can combine them together into another convergent sequence.
– A.S.
Commented Feb 23, 2016 at 14:03
• $\sum_{n \le N} a_{2n} \le \sum_{n \le 2N} a_{n}$ so $\sum_n a_{2n}$ converges Commented Feb 25, 2016 at 7:53

HINT: any absolutely convergent series is convergent.Let me know if you want the full solution

• That is another theorem to prove. Commented Feb 23, 2016 at 14:05
• This would be the main way to do this, but if you don't know the theorem try with A.S hint, it works too!
– Nick
Commented Feb 23, 2016 at 14:08
• Both has come to my mind but I was under the impression that there is one very easy answer. Commented Feb 23, 2016 at 14:09
• Yes, as many times it's easy if you know the right theorem. Try looking on your book for mine, you'll study it for sure anyway!
– Nick
Commented Feb 23, 2016 at 14:11

Theorem: If $\sum |a_n|$ is convergent, then $\sum a_n$ is convergent.

Proof: Let $b_n=|a_n|+a_n \Rightarrow b_n=0 \: \: \mbox{or} \: \: b_n=2|a_n| \Rightarrow 0\le b_n \le 2|a_n| \stackrel{\mbox{DCT}}{\Rightarrow} \sum b_n$ is convergent. Therefore, $\sum b_n - |a_n|=\sum a_n$ is convergent.