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Question Number 183690 by paul2222 last updated on 28/Dec/22

Answered by mr W last updated on 29/Dec/22

$$T_n=\frac{n}{a_n}$$$$a_n=1^6+2^6+3^6+...+n^6$$$$=\frac{n(n+1)(2n+1)(3n^4+6n^3-3n+1)}{42}$$$$T_n=\frac{42}{(n+1)(2n+1)(3n^4+6n^3-3n+1)}$$$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{42}{(n+1)(2n+1)(3n^4+6n^3-3n+1)}$$$$=42\underset{n=1}{\sum ^{\mathrm{\infty }}}[\frac{45n^3+21n^2-57n+30}{31(3n^4+6n^3-3n+1)}+\frac{32}{31(2n+1)}-\frac{1}{n+1}]$$$$.....$$

Commented by paul2222 last updated on 30/Dec/22

$$\mathit{W}ow$$

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