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.

  • $\begingroup$ My bad, it shows that that the original sum cannot be absolutely convergent. $\endgroup$ Oct 17 '16 at 0:36
  • $\begingroup$ Note that conditionally convergent series can be made to diverge themselves. $\endgroup$ Oct 17 '16 at 0:39
  • $\begingroup$ @SimpleArt, this is a good observation. However, the theorem supporting such reasoning is not presented until a later section in my textbook, meaning I don't have the tools of absolute and conditional convergence just yet. $\endgroup$
    – Decaf-Math
    Oct 17 '16 at 0:45
  • $\begingroup$ You change $\frac{1}{\sqrt{n}}$ to $\frac{1}{n}$ at one point up there. $\endgroup$
    – user361424
    Oct 17 '16 at 0:47
  • 2
    $\begingroup$ The definition of $\sum_{n=1}^\infty a_n$ converges is that the sequence $A(k) = \sum_{n=1}^k a_n$ converges. Here you have $A(k) = B(k)+C(k)$ where $B(k) = \sum_{n=1}^k \frac{(-1)^{n+1}}{\sqrt{n}}$ and $C(k) = \sum_{n=1}^k\frac{1}{k}$. As you said $B(k)$ converges and $C(k)$ diverges, therefore $A(k)$ diverges $\endgroup$
    – reuns
    Oct 17 '16 at 0:48

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.


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