Question Number 65981 by Tanmay chaudhury last updated on 07/Aug/19
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Answered by jimful last updated on 07/Aug/19
![let s_n =Σ1/n. Σ(k+1)/k •Σk/(k+1) =(n+s_n )(n−s_(n+1) +1) n^2 −n/(n+1)−s_n s_(n+1) +n+s_n =p(n) let q(n)=s_n s_(n+1) −s_n +n/(n+1) and r(n)=(1/8)n^2 −n we know that q(4)=((25)/(12))∙(s_5 −1)+(4/5)>2 and q(n) is incresing. since r(n) has mininum value −2 when n=4, (9/8)n^2 −p(n)=r(n)+q(n)≥0 for n≥4. q(1)+r(1)=1−7/8 q(2)+r(2)=23/12−3/2 q(3)+r(3)=8/72 ........](https://www.tinkutara.com/question/Q66011.png)
$${let}\:{s}_{{n}} =\Sigma\mathrm{1}/{n}. \\ $$$$\Sigma\left({k}+\mathrm{1}\right)/{k}\:\bullet\Sigma{k}/\left({k}+\mathrm{1}\right) \\ $$$$=\left({n}+{s}_{{n}} \right)\left({n}−{s}_{{n}+\mathrm{1}} +\mathrm{1}\right) \\ $$$${n}^{\mathrm{2}} −{n}/\left({n}+\mathrm{1}\right)−{s}_{{n}} {s}_{{n}+\mathrm{1}} +{n}+{s}_{{n}} ={p}\left({n}\right) \\ $$$$ \\ $$$${let}\:{q}\left({n}\right)={s}_{{n}} {s}_{{n}+\mathrm{1}} −{s}_{{n}} +{n}/\left({n}+\mathrm{1}\right) \\ $$$${and}\:{r}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{8}}{n}^{\mathrm{2}} −{n} \\ $$$${we}\:{know}\:{that}\:{q}\left(\mathrm{4}\right)=\frac{\mathrm{25}}{\mathrm{12}}\centerdot\left({s}_{\mathrm{5}} −\mathrm{1}\right)+\frac{\mathrm{4}}{\mathrm{5}}>\mathrm{2} \\ $$$${and}\:{q}\left({n}\right)\:{is}\:{incresing}. \\ $$$${since}\:{r}\left({n}\right)\:{has}\:{mininum}\:{value}\:−\mathrm{2}\:{when}\:{n}=\mathrm{4}, \\ $$$$\frac{\mathrm{9}}{\mathrm{8}}{n}^{\mathrm{2}} −{p}\left({n}\right)={r}\left({n}\right)+{q}\left({n}\right)\geqslant\mathrm{0}\:{for}\:{n}\geqslant\mathrm{4}. \\ $$$${q}\left(\mathrm{1}\right)+{r}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{7}/\mathrm{8} \\ $$$${q}\left(\mathrm{2}\right)+{r}\left(\mathrm{2}\right)=\mathrm{23}/\mathrm{12}−\mathrm{3}/\mathrm{2} \\ $$$${q}\left(\mathrm{3}\right)+{r}\left(\mathrm{3}\right)=\mathrm{8}/\mathrm{72}\:…….. \\ $$
Commented by Tanmay chaudhury last updated on 07/Aug/19
![thank you sir](https://www.tinkutara.com/question/Q66015.png)
$${thank}\:{you}\:{sir} \\ $$