Question Number 61804 by maxmathsup by imad last updated on 09/Jun/19
$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\sum_{{k}=\mathrm{2}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{k}^{{n}} \:\:{k}!} \\ $$
Commented by maxmathsup by imad last updated on 15/Jun/19
![S =Σ_(k=2) ^∞ (1/(k!)) Σ_(n=2) ^∞ ((1/k))^n =Σ_(k=2) ^∞ (w_k /(k!)) w_k =Σ_(n=2) ^∞ ((1/k))^n =Σ_(n=0) ^∞ ((1/k))^n −1−(1/k) =(1/(1−(1/k))) −1−(1/k) =(k/(k−1)) −1−(1/k) =((k−1 +1)/(k−1)) −1−(1/k) =(1/(k−1)) −(1/k) ⇒ S =Σ_(k=2) ^∞ (1/(k!))((1/(k−1)) −(1/k)) =Σ_(k=2) ^∞ (1/((k−1)k!)) −Σ_(k=2) ^∞ (1/(k k!)) =Σ_(k=1) ^∞ (1/(k(k+1)!)) −Σ_(k=2) ^∞ (1/(kk!)) let find Σ_(k=2) ^∞ (1/(kk!)) =Σ_(k=1) ^∞ (1/(kk!)) −1 we have e^x =Σ_(n=0) ^∞ (x^k /(k!)) ⇒ ∫_0 ^x e^t dt = Σ_(k=0) ^∞ [ (t^(k+1) /((k+1)k!))]_0 ^x +c ⇒ e^x −1 =Σ_(k=0) ^∞ (x^(k+1) /((k+1)k!)) x=1 ⇒e−1 =Σ_(k=0) ^∞ (1/((k+1)k!)) Σ_(k=1) ^∞ (1/(k(k+1)!)) =Σ_(k=1) ^∞ (1/(k(k+1)k!)) =Σ_(k=1) ^∞ ((1/k)−(1/(k+1)))(1/(k!)) =Σ_(k=1) ^∞ (1/(kk!)) −Σ_(k=1) ^∞ (1/((k+1)k!)) ⇒ S =Σ_(k=1) ^∞ (1/(kk!)) −Σ_(k=1) ^∞ (1/((k+1)k!)) −Σ_(k=2) ^∞ (1/(kk!)) = 1−{ Σ_(k=0) ^∞ (1/((k+1)k!)) −1} =1−{e−1−1} =1−e+2 =3−e ⇒ S =3−e .](https://www.tinkutara.com/question/Q62071.png)
$${S}\:=\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{k}!}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\left(\frac{\mathrm{1}}{{k}}\right)^{{n}} \:=\sum_{{k}=\mathrm{2}} ^{\infty} \:\:\frac{{w}_{{k}} }{{k}!} \\ $$$${w}_{{k}} =\sum_{{n}=\mathrm{2}} ^{\infty} \:\left(\frac{\mathrm{1}}{{k}}\right)^{{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{k}}\right)^{{n}} \:−\mathrm{1}−\frac{\mathrm{1}}{{k}}\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{k}}}\:−\mathrm{1}−\frac{\mathrm{1}}{{k}}\:=\frac{{k}}{{k}−\mathrm{1}}\:−\mathrm{1}−\frac{\mathrm{1}}{{k}} \\ $$$$=\frac{{k}−\mathrm{1}\:+\mathrm{1}}{{k}−\mathrm{1}}\:−\mathrm{1}−\frac{\mathrm{1}}{{k}}\:=\frac{\mathrm{1}}{{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{{k}}\:\Rightarrow\:{S}\:=\sum_{{k}=\mathrm{2}} ^{\infty} \frac{\mathrm{1}}{{k}!}\left(\frac{\mathrm{1}}{{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{{k}}\right) \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right){k}!}\:−\sum_{{k}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{{k}\:{k}!} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)!}\:−\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{kk}!}\:\:{let}\:{find}\:\:\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{kk}!}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{kk}!}\:−\mathrm{1} \\ $$$${we}\:{have}\:{e}^{{x}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{k}} }{{k}!}\:\Rightarrow\:\int_{\mathrm{0}} ^{{x}} \:{e}^{{t}} {dt}\:=\:\sum_{{k}=\mathrm{0}} ^{\infty} \left[\:\frac{{t}^{{k}+\mathrm{1}} }{\left({k}+\mathrm{1}\right){k}!}\right]_{\mathrm{0}} ^{{x}} \:+{c}\:\:\Rightarrow \\ $$$${e}^{{x}} −\mathrm{1}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{k}+\mathrm{1}} }{\left({k}+\mathrm{1}\right){k}!} \\ $$$${x}=\mathrm{1}\:\Rightarrow{e}−\mathrm{1}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){k}!} \\ $$$$\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)!}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right){k}!}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)\frac{\mathrm{1}}{{k}!}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{kk}!}\:−\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){k}!} \\ $$$$\Rightarrow\:{S}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{kk}!}\:−\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){k}!}\:−\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{kk}!} \\ $$$$=\:\mathrm{1}−\left\{\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){k}!}\:−\mathrm{1}\right\}\:=\mathrm{1}−\left\{{e}−\mathrm{1}−\mathrm{1}\right\}\:=\mathrm{1}−{e}+\mathrm{2}\:=\mathrm{3}−{e}\:\:\Rightarrow \\ $$$${S}\:=\mathrm{3}−{e}\:. \\ $$
Answered by perlman last updated on 09/Jun/19
$$\sum_{{n}=\mathrm{2}} ^{\infty} \sum_{{k}=\mathrm{2}} ^{\infty} \frac{\mathrm{1}}{{k}^{{n}} {k}!}=\sum_{{k}=\mathrm{2}} ^{+\infty} \sum_{{n}=\mathrm{2}} ^{+\infty} \frac{\mathrm{1}}{{k}^{{n}} {k}!} \\ $$$${justify} \\ $$$$\forall{k},{n}\geqslant\mathrm{2} \\ $$$$\sum_{{n}=\mathrm{2}} ^{+\infty} \sum_{{k}=\mathrm{2}} ^{+\infty} \frac{\mathrm{1}}{{k}!{k}^{{n}} }<\sum_{{n}} \frac{{e}}{\mathrm{2}^{{n}} }={e} \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \sum_{{k}=\mathrm{2}} ^{\infty} \frac{\mathrm{1}}{{k}^{{n}} {k}!}=\sum_{{k}=\mathrm{2}} ^{+\infty} \sum_{{n}=\mathrm{2}} ^{+\infty} \frac{\mathrm{1}}{{k}^{{n}} {k}!}=\sum_{{k}} \frac{\mathrm{1}}{{k}!}\sum_{{n}=\mathrm{2}} \frac{\mathrm{1}}{{k}^{{n}} } \\ $$$$=\sum_{{k}} \frac{\mathrm{1}}{{k}!}.\frac{\mathrm{1}}{{k}^{\mathrm{2}} }.\frac{{k}}{{k}−\mathrm{1}}=\sum_{{k}\geqslant\mathrm{2}} \frac{\mathrm{1}}{{k}!{k}\left({k}−\mathrm{1}\right)}=\sum_{{k}} \frac{{k}−\left({k}−\mathrm{1}\right)}{{k}!{k}\left({k}−\mathrm{1}\right)}=\sum_{{k}} \frac{\mathrm{1}}{{k}!\left({k}−\mathrm{1}\right)}−\frac{\mathrm{1}}{{k}!{k}} \\ $$$$=\sum_{{k}} \frac{{k}−\left({k}−\mathrm{1}\right)}{{k}!\left({k}−\mathrm{1}\right)}−\sum_{{k}} \frac{\mathrm{1}}{{kk}!} \\ $$$$=\sum_{{k}} \frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!\left({k}−\mathrm{1}\right)}−\sum_{{k}=\mathrm{2}} ^{+\infty} \frac{\mathrm{1}}{{k}!}−\Sigma\frac{\mathrm{1}}{{k}!{k}} \\ $$$$=\sum_{{k}=\mathrm{2}} ^{+\infty} \frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!\left({k}−\mathrm{1}\right)}−\sum_{{k}=\mathrm{2}} ^{+\infty} \frac{\mathrm{1}}{{k}!{k}}−\sum_{{k}=\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{{k}!}+\sum_{{k}=\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!\left(\mathrm{2}−\mathrm{1}\right)}−{e}+\mathrm{1}+\mathrm{1}=\mathrm{3}−{e} \\ $$$$ \\ $$$$ \\ $$