Does the Series \(\sum_{k=1}^{\infty}\frac{2+(-1)^k}{5^k}\) Converge?

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\sum_{k=1}^{\infty}\frac{2+\left(-1 \right)^{k}}{5^{k}}

Hello, I am trying to determine if this series converges or diverges. I have tried comparison test and d'Alembert's test but I was not successful

Can anyone suggest me a test learned in Calculus 2 that will work?

Thank you
 
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Break it into a sum of two series?
 
Since you're asking if there exists another test you could also use.

<br /> \lim_{n \to \infty}\sqrt[n]{|a_n|}&lt;1<br />

It converges absolutely if the limit is smaller than 1.
 
Thank you, I have figured it out. I have another question if you don't mind

\sum_{k=1}^{\infty}\frac{\left|\cos k \right|}{k^{3}}

It looks like a riemann series but I don't really know how to deal with trigonometric functions in series yet. What to do? Can anyone explain me?

Thank you
 
What values can |cos k| take? Perhaps consider a p-series
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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