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Question Number 77234 by TawaTawa last updated on 04/Jan/20

Answered by mr W last updated on 04/Jan/20

$$={\int }_0^1{x}^{\frac{1}{2}}{x}^{\frac{1}{2}\times \frac{1}{3}}{x}^{\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4}}{x}^{\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4}\times \frac{1}{5}}...dx$$$$={\int }_0^1{x}^{\frac{1}{1\times 2}+\frac{1}{1\times 2\times 3}+\frac{1}{1\times 2\times 3\times 4}+\frac{1}{1\times 2\times 3\times 4\times 5}+...}dx$$$$={\int }_0^1{x}^{\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}-2}dx$$$$={\int }_0^1{x}^{e-2}dx$$$$=\frac{1}{e-1}{\left[x^{e-1}\right]}_0^1$$$$=\frac{1}{e-1}$$

Commented by TawaTawa last updated on 04/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by TawaTawa last updated on 04/Jan/20

$$\mathrm{Sir},\mathrm{the}\mathrm{textbook}\mathrm{got}:\frac{1}{\mathrm{e}-1}.\mathrm{The}\mathrm{textbook}\mathrm{is}\mathrm{wrong}\mathrm{right}.$$$$\mathrm{If}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{mistake}\mathrm{in}\mathrm{your}\mathrm{work}\mathrm{sir}.$$

Commented by TawaTawa last updated on 04/Jan/20

$$\mathrm{And}\mathrm{sir},\mathrm{how}\mathrm{is}\mathrm{the}\mathrm{summation}{\mathrm{x}}^{\mathrm{e}-1}$$

Commented by TawaTawa last updated on 04/Jan/20

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}$$

Commented by TawaTawa last updated on 04/Jan/20

Commented by TawaTawa last updated on 04/Jan/20

$$\mathrm{Similar}\mathrm{question}\mathrm{sir}.$$

Commented by Smail last updated on 04/Jan/20

$$Thereisamistakeonthe{3}^{rd}line.$$$$e=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!},nstartsfrom0not1.$$$$Thus,theanswertothisquestionis\frac{1}{e-1}$$

Commented by TawaTawa last updated on 04/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by TawaTawa last updated on 04/Jan/20

$$\mathrm{Thanks}\mathrm{for}\mathrm{correcting}\mathrm{it}\mathrm{sir}\mathrm{Mrw}.$$

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