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

stropa0u

Answered question

2022-01-21

Prove the following inequality for every

n \geq 1

\sum_{k = 1}^{n} \frac{1}{k^{2} + 3 k + 1} \leq \frac{13}{20}

Answer & Explanation

rakije2v

Beginner2022-01-22Added 12 answers

Since $\frac{1}{k^{2} + 3 k + 1}$ is monotone decreasing for $k \geq 0$ , we have
$\sum_{k = 1}^{n} \frac{1}{k^{2} + 3 k + 1} \leq \frac{1}{5} + \frac{1}{11} + \int_{2}^{\infty} \frac{1}{k^{2} + 3 k + 1} d k$
$< \frac{1}{5} + \frac{1}{11} + \int_{2}^{\infty} \frac{1}{k^{2} + 2 k + 1} d k$
$= \frac{1}{5} + \frac{1}{11} + \frac{- 1}{k + 1} ∣_{2}^{\infty}$
$= \frac{1}{5} + \frac{1}{11} + \frac{1}{3}$ < $\frac{13}{20}$
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
$\sum_{k = 1}^{n} \frac{1}{k^{2} + 3 k + 1} \leq \frac{1}{5} + \int_{1}^{\infty} \frac{1}{k^{2} + 3 k + 1} d k$
2. That integral on the right looks mighty unpleasant; the denominator doesnt

Eleanor Shaffer

Beginner2022-01-23Added 16 answers

\sum_{k = 1}^{n} \frac{1}{k^{2} + 3 k + 1} \leq \sum_{k = 1}^{n} \frac{1}{k (k + 3)}

= \sum_{k = 1}^{n} \frac{1}{3} (\frac{1}{k} - \frac{1}{k + 3})

= \frac{1}{3} (1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{n + 1} - \frac{1}{n + 2} - \frac{1}{n + 3}) \leq \frac{13}{\frac{11}{6}}

= \frac{11}{18}

RizerMix

Expert2022-01-27Added 656 answers

This time i will appeal to the famous result of the Basel problem and get that:

\sum_{k = 1}^{\infty} \frac{1}{k^{2} + 3 k + 1} \leq \sum_{k = 1}^{\infty} \frac{1}{(k + 1)^{2}} = \frac{π^{2}}{6} - 1

But

\frac{π^{2}}{6} - 1 \leq \frac{13}{20}

The proof is complete.

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