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Question Number 119192 by help last updated on 22/Oct/20

Answered by Olaf last updated on 22/Oct/20

$$\mathrm{\forall }k\in {\mathbb{N}}^{\ast },k^2+1>k^2$$$$\mathrm{\forall }k\in {\mathbb{N}}^{\ast },{(k^2+1)}^3>k^6$$$$\mathrm{\forall }k\in {\mathbb{N}}^{\ast },\frac{1}{{(k^2+1)}^3}<\frac{1}{k^6}$$$$\mathrm{\forall }k\in {\mathbb{N}}^{\ast },\frac{k}{{(k^2+1)}^3}<\frac{1}{k^5}$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k}{{(k^2+1)}^3}<\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^5}$$$$\mathrm{And}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^5}\mathrm{converges}.$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^5}=\zeta \left(5\right)\approx 1,037$$$$\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k}{{(k^2+1)}^3}\mathrm{converges}.$$$$$$

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