I have been wondering if this infinite series converges $$\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}$$

I tried to put it in wolfram alpha but it says that the ratio test is inconclusive, but when I do the ratio test I get: Let first $a_k :=\frac{1}{(4+(-1)^k)^k}$, then we have that $\frac{1}{5^k}\le a_k \le \frac{1}{3^k}$. We also see from here that a_k is always positive and now we have $\frac{1}{5}\le\sqrt[k]{a_k}\le\frac{1}{3}<1, \forall k\in\mathbb{N}$. From that we should actually have that the series converges or am I missing something?

One more thing that I also noticed is that, if I use my inequality we can also have that $$\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}\le\sum_{k=1}^{\infty} \frac{1}{3^k}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$$

Does that mean it converges to $\frac{3}{2}$?

  • 2
    $\begingroup$ I'm sure it converges. Your inequality says it all $\endgroup$ – Yuriy S May 1 '16 at 20:03
  • $\begingroup$ Comparison test. $\endgroup$ – Robert Israel May 1 '16 at 20:03
  • 1
    $\begingroup$ It's approximately $\frac{5}{12}$ by numerical estimation $\endgroup$ – Yuriy S May 1 '16 at 20:04

Notice that the terms of this series are positive and we have

$$\frac{1}{(4+(-1)^k)^k}\le 3^{-k}$$ and the geometric series $\sum 3^{-k}$ is convergent. Use comparison to conclude.

  • $\begingroup$ Does it also mean it converges to $\frac{3}{2}$ like the series of $3^{-k}$ does? $\endgroup$ – HeatTheIce May 1 '16 at 20:11
  • $\begingroup$ No, but certainly the sum of this series is $<\frac32$. $\endgroup$ – user296113 May 1 '16 at 20:12
  • $\begingroup$ Do not forget that in addition to being bounded above by the $3^{-k}$ series (guaranteeing convergence), you are also bounded by $5^{-k}$ from below (which helps you narrow down your value). Therefore, you know the series converges to some value $x$ such that $5/4\leq x \leq 3/2$. $\endgroup$ – AmateurDotCounter May 1 '16 at 20:36
  • $\begingroup$ @LetEpsilonBeLessThanZero Note that $\sum_1^\infty 5^{-k} = 1/4$. $\endgroup$ – stochasticboy321 May 1 '16 at 20:38
  • $\begingroup$ @HeatTheIce To compute the sum, try splitting the series into two - one with the odd terms, and another with the even terms. You'll notice that both are convergent geometric series. $\endgroup$ – stochasticboy321 May 1 '16 at 20:38

Does that mean it converges to $\frac{3}{2}$?

Here is a closed form of the series:

$$ \sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}=\color{blue}{\frac5{12}}. $$

Proof. By the absolute convergence, one is allowed to write $$ \begin{align} \sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}&=\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^{2k})^{2k}}+\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^{2k-1})^{2k-1}} \\\\&=\sum_{k=1}^{\infty} \frac{1}{25^k}+3\sum_{k=1}^{\infty} \frac{1}{9^k} \\\\&=\frac1{24}+\frac38 \\\\&=\frac5{12}. \end{align} $$


Use the $\;k\,-$ th root test:


and thus the series converges (observe it is a positive series)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.