 stropa0u

2022-01-21

Prove the following inequality for every $n\ge 1$:
$\sum _{k=1}^{n}\frac{1}{{k}^{2}+3k+1}\le \frac{13}{20}$ rakije2v

Since $\frac{1}{{k}^{2}+3k+1}$ is monotone decreasing for $k\ge 0$, we have
$\sum _{k=1}^{n}\frac{1}{{k}^{2}+3k+1}\le \frac{1}{5}+\frac{1}{11}+{\int }_{2}^{\mathrm{\infty }}\frac{1}{{k}^{2}+3k+1}dk$
$<\frac{1}{5}+\frac{1}{11}+{\int }_{2}^{\mathrm{\infty }}\frac{1}{{k}^{2}+2k+1}dk$
$=\frac{1}{5}+\frac{1}{11}+\frac{-1}{k+1}{\mid }_{2}^{\mathrm{\infty }}$
$=\frac{1}{5}+\frac{1}{11}+\frac{1}{3}$<$\frac{13}{20}$
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
$\sum _{k=1}^{n}\frac{1}{{k}^{2}+3k+1}\le \frac{1}{5}+{\int }_{1}^{\mathrm{\infty }}\frac{1}{{k}^{2}+3k+1}dk$
2. That integral on the right looks mighty unpleasant; the denominator doesnt Eleanor Shaffer

$\sum _{k=1}^{n}\frac{1}{{k}^{2}+3k+1}\le \sum _{k=1}^{n}\frac{1}{k\left(k+3\right)}$
$=\sum _{k=1}^{n}\frac{1}{3}\left(\frac{1}{k}-\frac{1}{k+3}\right)$
$=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right)\le \frac{13}{\frac{11}{6}}$
$=\frac{11}{18}$ RizerMix

This time i will appeal to the famous result of the Basel problem and get that: $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}+3k+1}\le \sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(k+1{\right)}^{2}}=\frac{{\pi }^{2}}{6}-1$ But $\frac{{\pi }^{2}}{6}-1\le \frac{13}{20}$ The proof is complete.