# proof of comparison test

Assume $|a_{k}|\leq b_{k}$ for all $k>n$. Then we define

 $s_{k}:=\sum_{i=k}^{\infty}|a_{i}|$

and

 $t_{k}:=\sum_{i=k}^{\infty}b_{i}.$

Obviously $s_{k}\leq t_{k}$ for all $k>n$. Since by assumption $(t_{k})$ is http://planetmath.org/node/601convergent $(t_{k})$ is bounded and so is $(s_{k})$. Also $(s_{k})$ is monotonic and therefore . Therefore $\sum_{i=0}^{\infty}a_{i}$ is absolutely convergent.

Now assume $b_{k}\leq a_{k}$ for all $k>n$. If $\sum_{i=k}^{\infty}b_{i}$ is divergent then so is $\sum_{i=k}^{\infty}a_{i}$ because otherwise we could apply the test we just proved and show that $\sum_{i=0}^{\infty}b_{i}$ is convergent, which is is not by assumption.

Title proof of comparison test ProofOfComparisonTest 2013-03-22 13:22:06 2013-03-22 13:22:06 mathwizard (128) mathwizard (128) 4 mathwizard (128) Proof msc 40A05