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I was trying to find the value of x in the following equation.

$$x=\sum _{ i=0 }^{ \infty } \left( i+1 \right) \cdot \left( \frac { 5 }{ 6 } \right) ^{ i }$$

In a computer simulation, I found that $x=6^2$

I believe the following is also true, I just do not know how to prove it. $$p^2=\sum_{i=0}^{ \infty } (i+1)\dot\ (\frac{p-1}{p})^i$$

Could anyone tell me what theory I should look into to for this problem?

haqnatural
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Xin
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    The sum depends on $n$, so you will not get that. If you meant to sum to $\infty$, see http://math.stackexchange.com/q/30732/ – Jonas Meyer Jun 30 '16 at 06:34

3 Answers3

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The series

$$\sum_{n=1}^{\infty} nr^n$$

converges when $|r| < 1$ and is equal to $\dfrac{r}{(r-1)^2}$.

Here's a quick way to prove it:

$$\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n$$

Ignoring questions about justification, differentiate both sides (the right term-by-term) to get

$$\frac{1}{(1-r)^2} = \sum_{n=1}^{\infty} nr^{n-1}$$

and multiply by $r$ to get the result.

MT_
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$$x=\sum _{ i=0 }^{ \infty } \left( i+1 \right) \cdot \left( \frac { 5 }{ 6 } \right) ^{ i }$$ $$\frac{5x}{6}=\sum _{ i=0 }^{ \infty } \left( i+1 \right) \cdot \left( \frac { 5 }{ 6 } \right) ^{ i+1 }$$ $$x-\frac{5x}{6}=1+\sum_{i=1}^{\infty}\left( \frac { 5 }{ 6 } \right) ^{ i }$$ after this is basic G.P formula

avz2611
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A slightly more general identity is given by stars and bars: $$\frac{1}{(1-z)^{k+1}}=\sum_{n\geq 0}\binom{n+k}{n}z^n\tag{1} $$ for any $z$ such that $\|z\|<1$. Just take $k=1$ and $z=\frac{5}{6}$.

Kamala
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