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Question Number 90787 by Maclaurin Stickker last updated on 26/Apr/20

$$Findtheinfinitesum$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(2n-1)(2n+1)(2n+3)}$$

Commented by mathmax by abdo last updated on 26/Apr/20

$$letdecomposeF\left(x\right)=\frac{1}{(2x-1)(2x+1)(2x+3)}$$$$F\left(x\right)=\frac{a}{2x-1}+\frac{b}{2x+1}+\frac{c}{2x+3}$$$$a=(2x-1)F\left(x\right){\mid }_{x=\frac{1}{2}}=\frac{1}{\left(2\right)\left(4\right)}=\frac{1}{8}$$$$b=(2x+1)F\left(x\right){\mid }_{x=-\frac{1}{2}}=\frac{1}{(-2)\left(2\right)}=-\frac{1}{4}$$$$c=(2x+3)F\left(x\right){\mid }_{x=-\frac{3}{2}}=\frac{1}{(-4)(-2)}=\frac{1}{8}\Rightarrow$$$$F\left(x\right)=\frac{1}{8(2x-1)}-\frac{1}{4(2x+1)}+\frac{1}{8(2x+3)}let$$$$S_n=\sum _{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}\Rightarrow S_n=\mathrm{\Sigma }F\left(k\right)$$$$=\frac{1}{8}\sum _{k=1}^n\frac{1}{2k-1}-\frac{1}{4}\sum _{k=1}^n\frac{1}{2k+1}+\frac{1}{8}\sum _{k=1}^n\frac{1}{2k+3}wehave$$$$\sum _{k=1}^n\frac{1}{2k-1}=1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}$$$$=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{2n}$$$$=H_{2n}-\frac{1}{2}{H}_n$$$$\sum _{k=1}^n\frac{1}{2k+1}{=}_{k=j-1}\sum _{j=2}^{n+1}\frac{1}{2j-1}=\sum _{j=1}^{n+1}\frac{1}{2j-1}-1$$$$=H_{2n+2}-\frac{1}{2}{H}_{n+1}-1$$$$\sum _{k=1}^n\frac{1}{2k+3}{=}_{k=j-2}\sum _{j=3}^{n+2}\frac{1}{2j-1}=\sum _{j=1}^{n+2}\frac{1}{2j-1}-1-\frac{1}{3}$$$$=H_{2n+4}-\frac{1}{2}{H}_{n+2}-\frac{4}{3}\Rightarrow$$$$8S_n=H_{2n}-\frac{1}{2}{H}_n-2(H_{2n+2}-\frac{1}{2}{H}_{n+1}-1)+H_{2n+4}-\frac{1}{2}{H}_{n+2}-\frac{4}{3}$$$$=H_{2n}-\frac{1}{2}{H}_n-2H_{2n+2}+H_{n+1}+2+H_{2n+4}-\frac{1}{2}{H}_{n+2}-\frac{4}{3}$$$$\sim ln\left(2n\right)+\gamma -\frac{1}{2}ln\left(n\right)-\frac{\gamma }{2}-2ln(2n+2)-2\gamma +ln(n+1)+\gamma$$$$+ln(2n+4)+\gamma -\frac{1}{2}ln(n+2)-\frac{\gamma }{2}+\frac{2}{3}\Rightarrow$$$$8S_n\sim ln\left(2\right)+ln\left(n\right)-\frac{1}{2}ln\left(n\right)-2ln\left(2\right)-2ln\left(n\right)+ln\left(n\right)$$$$+ln\left(2\right)+ln\left(n\right)-\frac{1}{2}ln\left(n\right)+\frac{2}{3}\Rightarrow 8S_n\sim \frac{2}{3}\Rightarrow S_n\sim \frac{1}{12}$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{1}{12}$$

Commented by Maclaurin Stickker last updated on 26/Apr/20

$$Thankyou,Iappreciateyourwork.$$

Commented by abdomathmax last updated on 27/Apr/20

$$youarewelcome$$

Answered by ajfour last updated on 26/Apr/20

$$S=\frac{1}{4}\mathrm{\Sigma }\frac{(2n+3)-(2n-1)}{(2n-1)(2n+1)(2n+3)}$$$$=\frac{1}{4}\mathrm{\Sigma }\frac{1}{(2n-1)(2n+1)}-\frac{1}{4}\mathrm{\Sigma }\frac{1}{(2n+1)(2n+3)}$$$$=\frac{1}{8}\mathrm{\Sigma }\frac{1}{2n-1}-\frac{1}{8}\mathrm{\Sigma }\frac{1}{2n+1}-\frac{1}{8}\mathrm{\Sigma }\frac{1}{2n+1}+\frac{1}{8}\mathrm{\Sigma }\frac{1}{2n+3}$$$$8S=(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)-2(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)+(\frac{1}{5}+\frac{1}{7}+...)$$$$=1-\frac{1}{3}=\frac{2}{3}$$$$\Rightarrow S=\frac{1}{12}.$$$$(anychancethismaybecorrect?)$$

Commented by mathmax by abdo last updated on 26/Apr/20

$$correctsirajfour$$

Commented by Maclaurin Stickker last updated on 26/Apr/20

$$Perfect!$$

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