# Does the series $\sum_{n=0}^{\infty} \frac{(-2)^n}{1+2^n}$ converge?

Does the series $\sum_{n=0}^{\infty} \dfrac{(-2)^n}{1+2^n}$ converge?

I see that it doesn't converge absolutely, so I am checking conditional convergence now.

$$\sum_{n=0}^{\infty} \dfrac{(-2)^n}{1+2^n}=\sum_{n=0}^{\infty} \dfrac{(-1)^n\cdot2^n}{1+2^n}$$

$b_n=\dfrac{2^n}{1+2^n}$

Clearly $b_n$ is decreasing.

$\lim_{b\to\infty} b_n = 1\neq 0$. What does this mean? I was told that the alternating series test can only test convergence. Once of the hypothesis for convergence failed, so does this mean my series diverges, or that I have to apply another test?

• You have to use another test. This does not nessecarily mean it diverges.
– user370967
Apr 15 '17 at 21:09
• $a_n \to 0$ is a necessary condition for the convergence of $\sum a_n$. Apr 15 '17 at 21:11
• Which other test could I use for this case? If I use another test then I have to factor in the $(-1)^n$ into the problem Apr 15 '17 at 21:11
• $n$-th term test... as @DanielFischer says... Apr 15 '17 at 22:03

You have a problem here in that $$\lim_{n \to \infty}\sum _{k=1}^{2n}\frac{\left(-2\right)^k}{1+2^k}\approx0.205699255537$$ $$\lim_{n \to \infty}\sum _{k=1}^{2n+1}\frac{\left(-2\right)^k}{1+2^k}\approx -0.794300744463$$ For sufficiently large $n$ we are basically just alternating between these values as we increment $n$, so your series does not converge. Of course, this is overkill; all we have to do is note that $\displaystyle\lim_{n \to \infty}\frac{(-2)^n}{1+2^n}\neq0$ which means your series does not converge.
• Yes but I have the evaluate the limit in order to deduce that it doesnt go to $0$ Apr 15 '17 at 21:53
• We can use Leibniz criterion for the alternating series test. $$b_{n+1} - b_n = -\frac{2^n}{(1+2^n)(1+2^{n+1})} < 0.$$ So the alternating series is monotonically decreasing, yet its limit $b_n\to 1$ as $n\to\infty$. Thus, the series does not converge. Apr 15 '17 at 22:02