Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 145200 by qaz last updated on 03/Jul/21

$$\mathrm{evaluate}::\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}!({\mathrm{n}}^4+{\mathrm{n}}^2+1)}=\frac{\mathrm{e}}{2}$$

Answered by mindispower last updated on 03/Jul/21

$$n^4+n^2+1=n^4+2n^2+1-n^2={(n^2+1)}^2-n^2$$$$=(n^2+n+1)(n^2-n+1)$$$$\frac{1}{(n^2+n+1)(n^2-n+1)}=\frac{1}{2n}(.\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1})$$$$S=\sum _{n⩾0}\frac{1}{n!(n^4+n^2+1)}\sum _{n⩾0}f\left(n\right)=1+\sum _{n⩾1}f\left(n\right)$$$$=1+\sum _{n⩾1}\frac{1}{2n\left(n\right)!}.(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1})=S$$$$\sum _{n⩾1}\frac{1}{\left(2n\right)\left(n\right)!}.\frac{1}{n^2-n+1}$$$$=\frac{1}{2}+\sum _{n⩾1}\frac{1}{2(n+1).(n+1)!}.\frac{1}{(n^2+n+1)}$$$$S=\frac{3}{2}+\sum _{n⩾1}\frac{1}{2{(n+1)}^{2}n!(n^2+n+1)}-\frac{1}{2n(n^2+n+1)n!}$$$$=\frac{3}{2}+\sum _{n⩾1}\frac{1}{(n^2+n+1)n!}(\frac{1}{2{(n+1)}^2}-\frac{1}{2n})$$$$\frac{3}{2}+\mathrm{\Sigma }\left(\frac{n-{(n+1)}^2}{2n{(n+1)}^2}\right).\frac{1}{n^2+n+1}.\frac{1}{n!}$$$$=\frac{3}{2}-\frac{1}{2}.\sum _{n⩾1}\frac{n^2+n+1}{n(n+1)}.\frac{1}{n!(n^2+n+1)}$$$$=\frac{3}{2}-\frac{1}{2}\sum _{n⩾1}\frac{1}{n(n+1)(n+1)!}$$$$\sum _{n⩾1}\frac{1}{n(n+1)(n+1)!}=\sum _{n⩾1}\frac{1}{n(n+1)!}-\frac{1}{(n+1)(n+1)!}$$$$\text{Missing left or extra right}$$$$-\sum _{n⩾1}\frac{1}{(n+1)!}$$$$=1-(e-2)=3-e$$$$S=\frac{3}{2}-\frac{1}{2}(3-e)=\frac{e}{2}$$$$$$$$$$$$$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com