proof of comparison test

Assume $|a_k|\le b_k$ for all $k>n$. Then we define

$$s_k:=\sum _{i=k}^{\mathrm{\infty }}|a_i|$$

and

$$t_k:=\sum _{i=k}^{\mathrm{\infty }}{b}_i.$$

Obviously $s_k\le t_k$ 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 ${\sum }_{i=0}^{\mathrm{\infty }}{a}_i$ is absolutely convergent.

Now assume $b_k\le a_k$ for all $k>n$. If ${\sum }_{i=k}^{\mathrm{\infty }}{b}_i$ is divergent then so is ${\sum }_{i=k}^{\mathrm{\infty }}{a}_i$ because otherwise we could apply the test we just proved and show that ${\sum }_{i=0}^{\mathrm{\infty }}{b}_i$ is convergent, which is is not by assumption.

Title	proof of comparison test
Canonical name	ProofOfComparisonTest
Date of creation	2013-03-22 13:22:06
Last modified on	2013-03-22 13:22:06
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	4
Author	mathwizard (128)
Entry type	Proof
Classification	msc 40A05