I have this one:
$$ \sum_{n=1}^{\infty} {\frac{\ln{(3n^{2}+5)}x^{n}}{n^2 - 3n +5}} $$
I tried with the classic method:
$$ \sum_{n=1}^{\infty} {\frac{\ln{(3n^{2}+5)}((n+1)^2 -3(n+1) + 5)x^{n}}{(n^2 - 3n +5)(\ln{3(n+1)^{2}+5})(x^{n+1})}} $$
But, I can't get to answer, someone can help me please?
Hint
For large values of $n$
$$ {\frac{\ln{(3n^{2}+5)}\Big((n+1)^2 -3(n+1) + 5\Big)}{(n^2 - 3n +5)\ln\Big({3(n+1)^{2}+5}\Big)}}$$ can be approximated by $$1+\frac{2-\frac{2}{\log (3)+2 \log(n) }}{n}+O\left(\left(\frac{1}{n}\right)^2\right)$$