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Let $\sum_{n=0}^{\infty}a_nx^n$ of the function $\frac{\ln(1+x^2)}{x^2}$, give the expression of $a_n$.

I am at a loss here, I can figure out what $a_n$ is if it is only $\ln(1+x^2)$ but that is as far as I get.

R.vW
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  • You should know that for any $t\in(-1,1)$, we have $\ln(1+t)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}t^n$ and use it . – Adren Apr 04 '17 at 06:17

1 Answers1

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From here we have $$\ln(1+x^2)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x^2)^n}n\qquad\text{for }-1<x<1$$ So $$\frac{\ln(1+x^2)}{x^2}=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2(n-1)}}n\qquad\text{for }-1<x<1$$

Ángel Mario Gallegos
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