Show that the following series converges to $1$. $$\sum_{k=1}^\infty k \left(\frac{1}{2}\right)^{k+1}$$ I can't seem to find any good formulas or tricks for a sum like this.
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Do you know of Taylor series? Once you differentiate the formula for a geometric sum it's easy. – Sean Roberson Mar 12 '18 at 23:12
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See also this question. – Dietrich Burde Mar 12 '18 at 23:13
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Look to plug in 1/2 for $x$ in the derivative of the power series $\sum _{n=0}^\infty x^n$ – Jonathan Mar 12 '18 at 23:13
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This is equal to $\frac14\sum_{k=1}^\infty k\left(\frac12\right)^{k-1}$. Now, use the fact that$$\left(\frac1{1-x}\right)'=(1+x+x^2+x^3+\cdots)'=1+2x+3x^2+\cdots$$when $|x|<1$.
José Carlos Santos
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$$\begin{align} S = \frac{1}{2}+&2\frac{1}{4}+3\frac{1}{8}+\dots\\ \frac{1}{2}S = &\frac{1}{4}+2\frac{1}{8}+3\frac{1}{16}+\dots\\ \frac{1}{2}S = &\frac{1}{4}+\frac{1}{8}+\dots=1\\ S=2 \end{align}$$ I got the third line by subtracting the second from the first. For the above to be valid, you have to prove the series converges.
saulspatz
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