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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 $\frac{3}{2}$?

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    $\begingroup$ 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
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    $\begingroup$ It's approximately $\frac{5}{12}$ by numerical estimation $\endgroup$ – Yuriy S May 1 '16 at 20:04
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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.

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  • $\begingroup$ 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
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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 $$ \begin{align} \sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}&=\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^{2k})^{2k}}+\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^{2k-1})^{2k-1}} \\\\&=\sum_{k=1}^{\infty} \frac{1}{25^k}+3\sum_{k=1}^{\infty} \frac{1}{9^k} \\\\&=\frac1{24}+\frac38 \\\\&=\frac5{12}. \end{align} $$

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

$$\lim\sup_{k\to\infty}\sqrt[k]{\frac1{4+(-1)^k)^k}}=\lim\sup_{k\to\infty}\frac1{4+(-1)^k}=\frac13<1$$

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

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