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Question Number 206881 by efronzo1 last updated on 29/Apr/24

$$\:\:\cancel{\underline{\underbrace{\boldsymbol{{x}}}}} \\ $$

Answered by Skabetix last updated on 29/Apr/24

Σ_(n=0) ^∞ (x^n /(n!))=e^x →here x = 1 so 2 + (1/(2!))+...=e

$$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{n}!}={e}^{{x}} \rightarrow{here}\:{x}\:=\:\mathrm{1}\:{so}\:\mathrm{2}\:+\:\frac{\mathrm{1}}{\mathrm{2}!}+…={e} \\ $$

Commented by MM42 last updated on 29/Apr/24

? e+1 if x=1 ⇒ 1+(1/(1!))+(1/(2!))+(1/(3!))+...=2+(1/(2!))+(1/(3!))+...=e^1 =e

$$?\:{e}+\mathrm{1}\:\:\:\: \\ $$$${if}\:\:{x}=\mathrm{1}\:\Rightarrow\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+…=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+…={e}^{\mathrm{1}} ={e}\:\: \\ $$

Answered by mathzup last updated on 29/Apr/24

s=1+Σ_(k=0) ^∞ (1/(k!)) we know e^x =Σ_(k=0) ^∞ (1/(k!)) ⇒e=Σ_(k=0) ^∞ (1/(k!))(take x=1) ⇒s=1+e

$${s}=\mathrm{1}+\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{k}!}\:\:{we}\:{know}\:{e}^{{x}} =\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{k}!} \\ $$$$\Rightarrow{e}=\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{k}!}\left({take}\:{x}=\mathrm{1}\right)\:\Rightarrow{s}=\mathrm{1}+{e} \\ $$

Commented by MM42 last updated on 29/Apr/24

1+Σ_o ^∞ (1/(k!))=1+1+1+(1/(2!))+(1/(3!))+.. #2+(1/(2!))+(1/(3!))+....

$$\mathrm{1}+\underset{{o}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}!}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+.. \\ $$$$#\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+…. \\ $$

Answered by MM42 last updated on 29/Apr/24

$$\left.{ans}\right)\:{e} \\ $$

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