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Question Number 65981 by Tanmay chaudhury last updated on 07/Aug/19

Answered by jimful last updated on 07/Aug/19

$$let{s}_n=\mathrm{\Sigma }1/n.$$$$\mathrm{\Sigma }(k+1)/k\bullet \mathrm{\Sigma }k/(k+1)$$$$=(n+s_n)(n-s_{n+1}+1)$$$$n^2-n/(n+1)-s_n{s}_{n+1}+n+s_n=p\left(n\right)$$$$$$$$letq\left(n\right)=s_n{s}_{n+1}-s_n+n/(n+1)$$$$andr\left(n\right)=\frac{1}{8}{n}^2-n$$$$weknowthatq\left(4\right)=\frac{25}{12}\cdot (s_5-1)+\frac{4}{5}>2$$$$andq\left(n\right)isincresing.$$$$sincer\left(n\right)hasmininumvalue-2whenn=4,$$$$\frac{9}{8}{n}^2-p\left(n\right)=r\left(n\right)+q\left(n\right)⩾0forn⩾4.$$$$q\left(1\right)+r\left(1\right)=1-7/8$$$$q\left(2\right)+r\left(2\right)=23/12-3/2$$$$q\left(3\right)+r\left(3\right)=8/72........$$

Commented by Tanmay chaudhury last updated on 07/Aug/19

$$thankyousir$$

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