If $a_{n}a_{n+1}0$ (for all n)and $\sum a_{n}$ converges, then $\sum(-1)^{n} a_{n}$ converges.

Question

If $a_{n}>a_{n+1}>0$ (for all n)and $\sum a_{n}$ converges, then $\sum(-1)^{n} a_{n}$ converges.

Quizplus · Accepted Answer

Given that $\sum a_{n}$ converges and $a_{n} > a_{n+1} > 0$ for all $n$, it means $a_{n}$ is a decreasing sequence of positive terms converging to 0. By the Alternating Series Test, if the absolute values of the terms of an alternating series decrease monotonically to 0, then the series converges. Therefore, $\sum(-1)^{n} a_{n}$ converges.

If an>an+1>0a_{n}>a_{n+1}>0an​>an+1​>0 (For All N)and ∑an\sum a_{n}∑an​ Converges, Then ∑(−1)nan\sum(-1)^{n} a_{n}∑(−1)nan​ Converges

If $a_{n}>a_{n+1}>0$ (For All N)and $\sum a_{n}$ Converges, Then $\sum(-1)^{n} a_{n}$ Converges