# Prove the series is diverge

• Mar 29th 2010, 05:58 PM
wopashui
Prove the series is diverge
Show that if the series $\sum a_k$ converges and the series $\sum b_k$ diverges, then the series $\sum (a_k+b_k)$ diverges.

Give examples to show that if $\sum a_k$ and $\sum b_k$ both diverge, then each of the series

$\sum (a_k + b_k)$ and $\sum (a_k - b_k)$

may converge or may diverge
• Mar 29th 2010, 07:37 PM
Tinyboss
Quote:

Originally Posted by wopashui
Show that if the series $\sum a_k$ converges and the series $\sum b_k$ diverges, then the series $\sum (a_k+b_k)$ diverges.

Suppose $\sum (a_k+b_k)$ converges to L and $\sum a_k$ converges to K. For any $\varepsilon>0$, there exist integers N, M so that $\left|\sum_{k=1}^n(a_k+b_k)-L\right|<\varepsilon/2$ whenever n>N and $\left|\sum_{k=1}^na_k-K\right|<\varepsilon/2$ whenever n>M. Then since $|x-y|\le|x|+|y|$, we have $\left|\sum_{k=1}^nb_k-(L-K)\right|<\varepsilon$ whenever n>max(N,M), contradicting the divergence of that series.

Quote:

Originally Posted by wopashui
Give examples to show that if $\sum a_k$ and $\sum b_k$ both diverge, then each of the series

$\sum (a_k + b_k)$ and $\sum (a_k - b_k)$

may converge or may diverge

Just define $b_k=a_k=1$ to make $\sum (a_k + b_k)$ diverge and $\sum (a_k - b_k)$ converge, and $b_k=-a_k=1$ to get it the other way around.