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How do I evaluate the power series $$\sum_{n = 1}^{\infty}\dfrac{n}{9^{n}}$$ without using the formula for infinite geometric series?

I am interested in an approach which doesn't make use of infinite geometric series.

Any help would be highly appreciated. New approaches are most welcomed. Thanks.

BooleanCoder
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4 Answers4

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I'm not sure about the exact value, but you can compare it to the integral, which converges since $|\frac{1}{9}|<1$, $\int_{0}^{\infty}x a^x dx$, which will be of the same order, and then derive the difference using Bernoulli polynomials.

Alex
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Check with induction that the partial sum is $$\sum_{k=1}^N \frac{k}{9^k} = \frac{9 - 9^{-N}(8N+9)}{64}$$ Evaluate the limit when $N$ goes to infinity and you get the result.

Gabrielek
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  • This approach rather seems to be very naive – BooleanCoder Mar 22 '21 at 11:44
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    Just to be honest, as @DavidMitra pointed out in the reference, there is a known formula for partial sums of series like this, and this result is derived using "geometric series results". However you could pretend to be blind and use induction on a formula that comes from heaven and you are done lol. – Gabrielek Mar 22 '21 at 11:47
  • lmao Thanks anyways – BooleanCoder Mar 22 '21 at 11:49
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Notice that $x \frac{d}{dx} \frac{1}{1-x} = \sum_{i=0}^{\infty}ix^i $. So differenciating you get $$x \frac{d}{dx} \frac{1}{1-x} = \frac{x}{(1-x)^2}$$ Substituting $x= \frac{1}{9}$ you get required answer.

Strange
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    How is this not using the geometric series? Don't you need to know in advance that $\frac 1{1-x}=\sum_{i\geq 0} x^i$? – Giulio R Mar 22 '21 at 11:48
  • Yeah this does uses geometric series :( – Permutator Mar 22 '21 at 11:51
  • Well you can compute derivatives in zero...however not complete, but in general I do not see why one should restrain from using a basic identity such as the geometric series to use more involved methods (you also need to prove that your convergence radius of the Taylor expansion is greater than $1/9$ which involves study of analytic functions...) – Giulio R Mar 22 '21 at 11:53
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Denote $A=\sum_{n\geq 1} n9^{-n}$, $B=\sum_{n\geq 2} n9^{-n}$, $C=\sum_{n\geq 1} 9^{-n}$, $D=\sum_{n\geq 2} 9^{-n}$. You have the closed system $$ A=B+\frac 19,\quad 9B-A=C,\quad C=D+\frac 19,\quad 9D=C, $$ which gives the value of $A$. Of course, as a byproduct of the computation one also gets the value of the geometric series $C$.

Giulio R
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