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I'm supposed to calculate the sum of the power series $f(x)=\sum_{n=0}^{\infty}\frac{1}{4n^2-1}x^{2n+1}$.

I do the following rewrite: $$f(x)=\sum_{n=0}^{\infty}\frac{1}{4n^2-1}x^{2n+1}=\sum_{n=0}^{\infty}\frac{1}{(2n-1)(2n+1)}x^{2n+1}$$ Then I derive it and get $$f'(x)=\frac{d}{dx}\sum_{n=0}^{\infty}\frac{1}{(2n-1)(2n+1)}x^{2n+1}=\sum_{n=0}^{\infty}\frac{1}{2n-1}x^{2n}.$$ At this point I'm a little stuck. I've tried doing this in order to maybe derive another time: $$f'(x)=x\sum_{n=0}^{\infty}\frac{1}{2n-1}x^{2n-1}.$$ Deriving another time gives me: $$f''(x)=\sum_{n=0}^{\infty}\frac{1}{2n-1}x^{2n-1}+x\sum_{n=0}^{\infty}x^{2n-2}=\sum_{n=0}^{\infty}\frac{1}{2n-1}x^{2n-1}+\frac{1}{x}\sum_{n=0}^{\infty}x^{2n}=\sum_{n=0}^{\infty}\frac{1}{2n-1}x^{2n-1}+\frac{1}{x\cdot(1-x^2)}$$

From here I could use some partial fraction decomposition on the second term in f''(x) and play around with the first term (the sum) in f''(x) to get f(x), but it seems to need lots of work. I'm wondering if there might be an easier way that I'm not aware of. Thankful for some tips or guidance! :)

Jonathan

Thomas Andrews
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Jonathan Axelsson
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  • Maybe you can factor the denominator into $(2n-1)(2n+1)$ and use partial fractions? – Тyma Gaidash Dec 23 '21 at 00:04
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    Differentiate the identity again only after you divide it by $x$, in other words $(f'(x)/x)'=\left(\sum_{n=0}^{\infty}\frac 1{2n-1}\cdot x^{2n-1}\right)'$ – Pavel R. Dec 23 '21 at 00:11

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Recall the following Taylor expansions: $$\begin{align*} \log (1+x) &= \sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k} \\ -\log(1-x) &= \sum_{k=1}^\infty \frac{x^k}{k} \end{align*}$$

Adding them together you are left with two copies of each term with odd powers: $$\begin{align*} \log(1+x) - \log (1-x) &= 2\left( x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots \right) \\ &= 2 \sum_{k=1}^\infty \frac{x^{2k-1}}{2k-1} \\ &= 2x^{-1} \sum_{k=1}^\infty \frac{x^{2k}}{2k-1} \\ &= 2\frac{(f^\prime (x) + 1)}{x} \end{align*}$$

Can you take it from here?

Jose Avilez
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