# Divergence of positive series implies divergence of alternating series?

Consider the alternating series $\Sigma_{n=0}^{\infty} \left(\frac{2n+2}{2n+1}\right)^n (-1)^n$. The question is to assess its absolute convergence.

If I set $u_n = \left(\frac{2n+2}{2n+1}\right)^n = \left(1 + \frac{1}{2n+1} \right)^n$, then $\lim_{n\rightarrow \infty}u_n = e^{1/2}$, so I can only conclude that the alternating series does not converge absolutely. Yet my teacher's feedback is that this is enough to conclude that the alternating series diverges. I don't get her point. Am I missing something?

• The added terms don't go to zero if $n\to\infty$, so clearly, this series can't be converging! – user408856 May 27 '18 at 9:42

## 2 Answers

My guess is that what your teacher told you is that it is enough to note that we don't have $\lim_{n\to\infty} \left(\frac{2n+2}{2n+1}\right)^n (-1)^n=0$. And, yes, that's enough to conclude that the series is not absolutely convergent.

• I'd remove “absolutely”. – egreg May 27 '18 at 9:53

A series can converge only when its n-th term tends to 0 as $n \to \infty$. In this case $(-1)^{n}u_n$ does not tend to 0 so the series is not convergent, hence not absolutely convergent.