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The sum is as follows:

$$ \sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1}\\ $$

This is how I started:

$$ = \frac{1}{6}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n-1} \\ = \frac{1}{5}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n}\\\\ = \frac{1}{5}S\\ S = \frac{5}{6} + 2\left (\frac{5}{6}\right)^2 + 3\left (\frac{5}{6}\right)^3 + ... $$

I don't know how to group these in to partial sums and get the result. I also tried considering it as a finite sum (sum from 1 to n) and applying the limit, but that it didn't get me anywhere!

PS: I am not looking for the calculus method.


I tried to do it directly in the form of the accepted answer,

$$ \textrm{if} \ x= \frac{5}{6},\\ S = x + 2x^2 + 3x^3 + ...\\ Sx = x^2 + 2x^3 + 3x^4 + ...\\ S(1-x) = x + x^2 + x^3 + ...\\ \textrm{for x < 1},\ \ \sum_{n=1}^{\infty}x^n = -\frac{x}{x-1}\ (\textrm{I looked up this eqn})\\ S = \frac{x}{(1-x)^2}\\ \therefore S = 30\\ \textrm{Hence the sum} \sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1} = \frac{30}{5} = 6 $$

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2 Answers 2

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Letting $a = d = 1/6$ and $r = 5/6$, our sum is: $$ S = a + (a + d)r + (a + 2d)r^2 + (a + 3d)r^3 + \cdots $$ Scaling by $r$, we find that: $$ rS = ar + (a + d)r^2 + (a + 2d)r^3 + \cdots $$ Subtracting the two equations (by collecting like powers of $r$), we obtain: $$ (1 - r)S = a + dr + dr^2 + dr^3 + \cdots = a + dr(1 + r + r^2 + \cdots) = a + \frac{dr}{1 - r} $$ Hence, we conclude that: $$ S = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2} = \frac{1/6}{1 - 5/6} + \frac{(1/6)(5/6)}{(1 - 5/6)^2} = \frac{1}{6 - 5} + \frac{(1)(5)}{(6 - 5)^2} = 6 $$

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  • $\begingroup$ I tried the problem in this manner, would you please take a look at it? Also, are the a, d, and r substitutions made by using standard equations or do they come by intuition? $\endgroup$
    – user41235
    Jan 19, 2015 at 15:14
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hint: differentiate the identity

$$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$

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    $\begingroup$ Quote: "PS: I am not looking for the calculus method." $\endgroup$
    – Did
    Jan 19, 2015 at 7:14
  • $\begingroup$ @Did Well, the other answer uses the sum of an infinite geometric series, which is also "calculus method". I can't see any way out of using some calculus here. $\endgroup$
    – Timbuc
    Jan 19, 2015 at 9:58
  • $\begingroup$ @Timbuc Of course one would need a precise definition of "calculus methods" here but it seems obvious that the OP is after methods such as the (discrete) recursion used in Adriano's answer, not after methods using differentiation such as this one. $\endgroup$
    – Did
    Jan 19, 2015 at 10:42
  • $\begingroup$ @Timbuc: There is no a real issue here! The whole thing is that I did not pay attention to the OP's Quote. But as I said to know the problem see here. $\endgroup$ Jan 19, 2015 at 10:46

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