Prove the following inequality for every n\geq1: \sum_{k=1}^n\frac{1}{k^2+3k+1}\leq\frac{13}{20}

stropa0u

stropa0u

Answered question

2022-01-21

Prove the following inequality for every n1:
k=1n1k2+3k+11320

Answer & Explanation

rakije2v

rakije2v

Beginner2022-01-22Added 12 answers

Since 1k2+3k+1 is monotone decreasing for k0, we have 
k=1n1k2+3k+115+111+21k2+3k+1dk 
<15+111+21k2+2k+1dk 
=15+111+1k+12 
=15+111+13<1320
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 
k=1n1k2+3k+115+11k2+3k+1dk 
2. That integral on the right looks mighty unpleasant; the denominator doesnt

Eleanor Shaffer

Eleanor Shaffer

Beginner2022-01-23Added 16 answers

k=1n1k2+3k+1k=1n1k(k+3)
=k=1n13(1k1k+3)
=13(1+12+131n+11n+21n+3)13116
=1118
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

This time i will appeal to the famous result of the Basel problem and get that: k=11k2+3k+1k=11(k+1)2=π261 But π2611320 The proof is complete.

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