Prove the following inequality for every n≥1: ∑k=1n1k2+3k+1≤1320

Since ${\displaystyle \frac{1}{k^2+3k+1}}$ is monotone decreasing for ${\displaystyle k\ge 0}$, we have 
${\displaystyle \sum _{k=1}^n\frac{1}{k^2+3k+1}\le \frac{1}{5}+\frac{1}{11}+{\int }_2^{\mathrm{\infty }}\frac{1}{k^2+3k+1}dk}$ 
${\displaystyle <\frac{1}{5}+\frac{1}{11}+{\int }_2^{\mathrm{\infty }}\frac{1}{k^2+2k+1}dk}$ 
${\displaystyle =\frac{1}{5}+\frac{1}{11}+\frac{-1}{k+1}{\mid }_2^{\mathrm{\infty }}}$ 
${\displaystyle =\frac{1}{5}+\frac{1}{11}+\frac{1}{3}}$<$\frac{13}{20}$
1. I remembered that monotonic series can be bounded by integrals, by thinking of the series as a right Riemann sum. This suggests I try the bound 
${\displaystyle \sum _{k=1}^n\frac{1}{k^2+3k+1}\le \frac{1}{5}+{\int }_1^{\mathrm{\infty }}\frac{1}{k^2+3k+1}dk}$ 
2. That integral on the right looks mighty unpleasant; the denominator doesnt