How Do You Solve the Series Sum n/2^(n-1)?

djuiceholder
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Homework Statement


so I have learned how to do different problems of series.
but there's this problem that I spent hours last night but could not come up with anything.
which is-


∑ n/2^(n-1)
n=1



Homework Equations



I have no idea

The Attempt at a Solution


so from there i was able to take out 2 and end up with: 2 n /(2^n)

i don't know how to do after that. however, i thought about doing (1/2)^n but then again it does not make sense to me.

help needed. i will appreciate that. thanks
 
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Why have you cross-posted this problem to this forum and the Precalc forum?
 
Something tells me that the ratio test could helo here.
 
If you want to show the series converges, the ratio test can help. If you want to find the sum, the ratio test can't help. Hint: this looks like it's related to the series 1+x+x^2+x^3+... Take a derivative. Now figure out what to put x equal to.
 
i was supposed to evaluate it. I don't know how though
 
i posted this problem on a different folder and i was warned because it was a wrong place to post. anyways, i think i took it off
 
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...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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