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Let $a>0$. For $|\frac{1}{a^x}|<1, \sum_{n=0}^{\infty} \frac{n\cdot log(a)}{a^{nx}}=\frac{d}{dx}\sum_{n=1}^{\infty} \frac{-1}{a^{nx}}=\frac{d}{dx}\frac{a^x}{1-a^x}=\frac{a^x\cdot log(a)}{(1-a^x)^2}$. Hence $ \sum_{n=1}^{\infty} \frac{n}{a^{nx}}=\frac{a^x}{(1-a^x)^2}$, which gives $ \sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$, for $|x|<1$.

I don't know how to justify taking the derivative since it's not a power series, but it seems to give me the right answer. I was wondering if this was the standard way for evaluating this series, or if there is another, more common way, possibly involving generating functions and a differential equation?

SihOASHoihd
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2 Answers2

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Using derivatives with power series is a useful way to evaluate these power series, when the value of $x$ lies in the interval of convergence. In the case of the series in question, the interval of convergence is $(-1,1)$. So when $x\in (-1,1)$ we have that $\sum\frac{d}{dx}=\frac{d}{dx}\sum$ (i.e. you can swap differentiating the series with taking the series of the differentiated terms).

For this series one could also do the following (for $x\in(-1,1)$): $$\sum_{n=1}^\infty nx^n=x\sum_{n=1}^\infty nx^{n-1}=x\sum_{n=1}^\infty\left(\frac{d}{dx}x^n\right)=x\frac{d}{dx}\left(\sum_{n=1}^\infty x^n\right)=x\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{x}{(1-x)^2}$$

Dave
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    Don't we need uniform convergence (more than simply pointwise convergence) to be able to interchange sum and derivative? https://math.stackexchange.com/questions/147869/interchanging-the-order-of-differentiation-and-summation – leonbloy Jul 25 '17 at 23:47
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    @leonbloy since this is a power series with radius of convergence $R=1$ centered around $0$, we know that it converges uniformly on any interval $[-r,r]$ with $0<r<R$. And on the interval $(-R,R)$ we have the ability to swap the sum and derivative. One could check that $R=1$ is indeed the radius of convergence, but checking uniform continuity is not really necessary. Check out the section on differentiation: https://en.wikipedia.org/wiki/Power_series – Dave Jul 26 '17 at 01:26
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Usually people use a simpler approach, $x\frac{d}{dx}\sum_{n=1}^\infty x^n$. As with your method, a series that is a function of $x$ is differentiated termwise with respect to $x$. As long as the original and differentiated series have finite terms and converge, this is legitimate even if the differentiated terms aren't a "standard" series expression for a function, such as a Taylor series in $x$.

J.G.
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