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Question Number 111973 by bobhans last updated on 05/Sep/20

(1/(5!!)) + ((1.3)/(7!!)) + ((1.3.5)/(9!!)) + ... + ((1.3.5.7....95)/(99!!)) =?

$$\frac{\mathrm{1}}{\mathrm{5}!!}\:+\:\frac{\mathrm{1}.\mathrm{3}}{\mathrm{7}!!}\:+\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{9}!!}\:+\:...\:+\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}....\mathrm{95}}{\mathrm{99}!!}\:=? \\ $$

Commented by Dwaipayan Shikari last updated on 05/Sep/20

1.3.5.7.9.11...n=(((2n)!)/(2.4.6.8.10..n))=(((2n)!)/(2^n .n!)) Σ_(n=1) ^(48) ((((2n)!)/(2^n .n!))/((2n+3)!!)) If (2n+3)!! becomes (2n+3)! then Σ_(n=1) ^(48) (((2n)!)/(2^n .n!(2n+3)(2n+2)(2n+1)(2n)!))=Σ_(n=1) ^(48) (1/(2^n .n!(2n+3)(2n+2)(2n+1)))

$$\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}.\mathrm{11}...{n}=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}.\mathrm{10}..{n}}=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} .{n}!} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{48}} {\sum}}\frac{\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} .{n}!}}{\left(\mathrm{2}{n}+\mathrm{3}\right)!!} \\ $$$${If}\:\left(\mathrm{2}{n}+\mathrm{3}\right)!!\:\:{becomes}\:\left(\mathrm{2}{n}+\mathrm{3}\right)!\:\:{then} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{48}} {\sum}}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} .{n}!\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{48}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} .{n}!\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 05/Sep/20

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