Infinity Summation problem (ANSWERED)

(Big) hint: you can rewrite $$ \frac{1}{4n^4+16n^3+23n^2+14n+3}= 2\left( \frac{1}{2n+1}-\frac{1}{2n+3}\right) - \frac{1}{(n+1)^2} $$ and now, you have the sum of a telescoping series and a convergent series (whose sum is known).

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That is, we can compute the sum of the telescopic series as $$ \sum_{n=1}^N 2\left( \frac{1}{2n+1}-\frac{1}{2n+3}\right) = 2\left( \frac{1}{2\cdot 1+1}-\frac{1}{2N+3}\right) \xrightarrow[N\to\infty]{} 2\left( \frac{1}{2\cdot 1+1}-0\right) = \frac{2}{3} $$ while the last term will give $$ \sum_{n=1}^\infty \frac{1}{(n+1)^2} =\sum_{n=2}^\infty \frac{1}{n^2} =\sum_{n=1}^\infty \frac{1}{n^2} - 1 = \frac{\pi^2}{6}-1 $$ so that $$ \sum_{n=1}^\infty \frac{1}{4n^4+16n^3+23n^2+14n+3} = \frac{2}{3}- (\frac{\pi^2}{6}-1) = \boxed{\frac{10-\pi^2}{6}\,.} $$

Too long for a comment.

Interesting (at least to me !) is the problem of the partial sums$$S_p=\sum_{n=1}^p \frac{1}{4n^4+16n^3+23n^2+14n+3}$$ Skipping the steps which are almost identical to @Clement C.'s ones,we can end with the nice $$S_p=\frac{10 p+9}{3(2 p+3)}+\psi ^{(1)}(p+2)-\frac{\pi ^2}{6}$$and using the asymptotic $$\psi ^{(1)}(q)=\frac{1}{q}+\frac{1}{2 q^2}+\frac{1}{6 q^3}-\frac{1}{30 q^5}+O\left(\frac{1}{q^7}\right)$$ Continuing with Taylor series, we end with $$S_p=\frac{10-\pi^2}{6}-\frac 1 {12p^3}+\frac 8 {3p^4}+O\left(\frac{1}{p^5}\right)$$