Show that $\displaystyle \sum_{n=1}^\infty (-1)^{n+1}a_n$ diverges, where $\displaystyle a_n = {1 \over \sqrt n} + {(-1)^{n-1}\over n}$.
I just have a simple question in regards to divergence of a series. Here's what I did so far:
$$\begin{align}\sum_{n=1}^\infty (-1)^{n+1}a_n & = \sum_{n=1}^\infty \left[{(-1)^{n+1} \over \sqrt n} + {1 \over n}\right] \\ &=\underset{\text{converges by AST}}{\sum_{n=1}^\infty {(-1)^{n+1} \over \sqrt n}} + \underset{\text{diverges}}{\sum_{n=1}^\infty {1 \over n}} \end{align}$$
My question: Does it suffice to say that if one series diverges and another converges, that the entire series will diverge? Obviously we can't immediately invoke the AST at the beginning because the sequence $\{a_n\}$ is not nonincreasing, so my thought was to do some algebra to show that the sequence can be broken into both a convergent and divergent series.
Yes, if one converges and the other diverges, the sum diverges. If the sum converged, then the difference between the sum and the convergent term would converge, but this difference is the divergent term.