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 | ||
$$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{n}!}={e}^{{x}} \rightarrow{here}\:{x}\:=\:\mathrm{1}\:{so}\:\mathrm{2}\:+\:\frac{\mathrm{1}}{\mathrm{2}!}+...={e} \\ $$ | ||
Commented by MM42 last updated on 29/Apr/24 | ||
$$?\:{e}+\mathrm{1}\:\:\:\: \\ $$$${if}\:\:{x}=\mathrm{1}\:\Rightarrow\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+...=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+...={e}^{\mathrm{1}} ={e}\:\: \\ $$ | ||
Answered by mathzup last updated on 29/Apr/24 | ||
$${s}=\mathrm{1}+\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{k}!}\:\:{we}\:{know}\:{e}^{{x}} =\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{k}!} \\ $$$$\Rightarrow{e}=\sum_{{k}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{k}!}\left({take}\:{x}=\mathrm{1}\right)\:\Rightarrow{s}=\mathrm{1}+{e} \\ $$ | ||
Commented by MM42 last updated on 29/Apr/24 | ||
$$\mathrm{1}+\underset{{o}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}!}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+.. \\ $$$$#\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+.... \\ $$ | ||
Answered by MM42 last updated on 29/Apr/24 | ||
$$\left.{ans}\right)\:{e} \\ $$ | ||