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I'm trying to find the sum of the following series:

$$\sum^\infty_{n=1}\frac {(x-3)^{2n}}{2n}$$

I tried to "convert" it to a simple geometrical series, but with no luck. Has someone any idea?


Thanks for inspiration! My solution: $$\sum^\infty_{n=1}\frac {(x-3)^{2n}}{2n} = \sum^\infty_{n=1}{\frac{z^{2n}}{2n}}=\sum^\infty_{n=1}{\int_0^z{z^{2n-1}}dz}=\int^z_0{\sum^\infty_{n=1}{z^{2n-1}}}=\int^z_0{\frac{z}{1-z^2}}=-\frac12 log(-x^2+6x-8)$$

I was wondering if there are any conditions?

$x\neq4$ and $x\neq2$ and $(x-3)^2\leq1$

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  • $\begingroup$ @dimitrij Do you know the sum of the series $\sum\limits_{n\geqslant1}\frac{z^n}n$, when it exists? $\endgroup$
    – Did
    Apr 26, 2015 at 17:31
  • $\begingroup$ Hint Remember the mclaurin series of the logarithm. What happens then to your series when $(x-3)^2 < 1$ $\endgroup$ Apr 26, 2015 at 17:37
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    $\begingroup$ Note that the $n$-th term is $\int_0^z t^{2n-1}\,dt$. So our series is obtained by integrating the geometric series $\sum t^{2n-1}$ term by term. $\endgroup$ Apr 26, 2015 at 17:47
  • $\begingroup$ Thanks for inspiration! You guys are very smart and fast. $\endgroup$ Apr 26, 2015 at 18:28
  • $\begingroup$ An $n$ in the denominator can signal we should integrate something, just like an $n$ in the numerator can indicate we should differentiate something. $\endgroup$ Apr 26, 2015 at 18:34

2 Answers 2

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Hint: $$\log(1-x)=-\sum_{n=1}^{\infty }\frac{x^n}{n}$$

Let $x=x^2$, then: $$\log(1-x^2)=-\sum_{n=1}^{\infty }\frac{x^{2n}}{n}$$

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Since for any $|z|<1$ we have: $$ -\log(1-z)=\sum_{n\geq 1}\frac{z^n}{n}, \tag{1} $$ $$ -\log(1+z)=\sum_{n\geq 1}\frac{(-1)^n\, z^n}{n}, \tag{2} $$ (you can check them both by integrating termwise the usual geometric series for $\frac{1}{1-z}$ or $\frac{1}{1+z}$) by just summing $(1)$ and $(2)$ one gets: $$ -\frac{1}{2}\log(1-z^2)=\sum_{n\geq 1}\frac{z^{2n}}{2n} \tag{3}$$ for any $|z|<1$. Now you just need to replace $z$ with $x-3$.

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