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.

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


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?

  • $\begingroup$ You have to use another test. This does not nessecarily mean it diverges. $\endgroup$ – user370967 Apr 15 '17 at 21:09
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    $\begingroup$ $a_n \to 0$ is a necessary condition for the convergence of $\sum a_n$. $\endgroup$ – Daniel Fischer Apr 15 '17 at 21:11
  • $\begingroup$ 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 $\endgroup$ – K Split X Apr 15 '17 at 21:11
  • $\begingroup$ $n$-th term test... as @DanielFischer says... $\endgroup$ – Simply Beautiful Art 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.

  • $\begingroup$ Would you simply compare even and odd values seperately and show that since they are not the same, then the series diverges? $\endgroup$ – K Split X Apr 15 '17 at 21:49
  • $\begingroup$ @KSplitX you could, but as I said... overkill. If the terms of a series don't go to zero, the series doesn't converge. That is the whole of your proof. $\endgroup$ – Brevan Ellefsen Apr 15 '17 at 21:50
  • $\begingroup$ Yes but I have the evaluate the limit in order to deduce that it doesnt go to $0$ $\endgroup$ – K Split X Apr 15 '17 at 21:53
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    $\begingroup$ 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. $\endgroup$ – Hoc Ngo Apr 15 '17 at 22:02

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