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

Beginner2022-01-22Added 12 answers

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

Beginner2022-01-23Added 16 answers

RizerMix

Skilled2022-01-27Added 437 answers

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