Summing the Series: $\sum (1+2^n)/3^n$

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Homework Statement


\sum (1+2^n)/3^n

Find its sum


Homework Equations





The Attempt at a Solution


I don't know how to figure this one out. I took the limit and it was zero so I know it converges. I don't know where to start on this one. The answer is 5/2.
 
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Write this as the sum of two summations.
 
More precisely, the sum of two geometric series.
 
And just because the the terms approach zero, the series always convergent. Think of the harmonic series.
 
Typo? I'm quite sure Gib Z meant "just because the terms approach zero, the series is not always convergent". Yes, this particular series converges but your statement " I took the limit (of the sequence of terms) and it was zero so I know it converges" does not follow.
The fact that the limit of the sequence of terms goes to zero is necessary for convergence but not sufficient.
 
Typo indeed :( Was in a rush for that post lol.
 
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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