# Thread: Prove the series is diverge

1. ## Prove the series is diverge

Show that if the series $\displaystyle \sum a_k$ converges and the series $\displaystyle \sum b_k$ diverges, then the series $\displaystyle \sum (a_k+b_k)$ diverges.

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

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

may converge or may diverge

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

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

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

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