Find the sum of the series: $$ \sum_ {n=1} ^{\infty} {nz^n} , $$ $$ |z| < 1$$
Where do I start from? Can I use the root test?
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1Can you sum $\sum_{n=1}^\infty z^n$? If so, notice that that your series is exactly $z \cdot \frac{d}{dz} (\sum_{n=1}^\infty z^n)$. – xyzzyz Apr 27 '16 at 23:17
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1The Root Test will tell you that the series converges, but it will not give you the value. – André Nicolas Apr 27 '16 at 23:22
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There was a page where they evaluated this first $$ \sum_ {n=1} ^{\infty} {nz^{n-1}} $$ but I could not find it now. They used a theorem to do that. – Josh Apr 27 '16 at 23:32
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This has been asked very, very many times before on this site. – Akiva Weinberger Apr 27 '16 at 23:42
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See here; your equation is that one times $z$. – Akiva Weinberger Apr 27 '16 at 23:44
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Here is a useful finite evaluation: $$ 1+z+z^2+...+z^n=\frac{1-z^{n+1}}{1-z}, \quad |z|<1. \tag1 $$ Then by differentiating $(1)$ you get $$ 1+2z+3z^2+...+nz^{n-1}=\frac{1-z^{n+1}}{(1-z)^2}+\frac{-(n+1)z^{n}}{1-z}, \quad |z|<1, \tag2 $$ and by making $n \to +\infty$ in $(2)$, using $|z|<1$, and multiplying by $z$ gives
$$ \sum_{n=1}^\infty nz^n=z+2z^2+3z^3+...+nz^n+...=\frac{z}{(1-z)^2}. \tag3 $$
Olivier Oloa
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$$\sum_ {n=1} ^{\infty} {nz^n} =\left(\sum_ {n=1} ^{\infty} {z^n} \right)^2$$
Rene Schipperus
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Yeah,yeah you guys love to criticise, dont you. Actually I didnt really pay any attention to the indices. – Rene Schipperus Apr 28 '16 at 04:25
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