Question-201091

Question Number 201091 by MrGHK last updated on 29/Nov/23

Commented by Frix last updated on 29/Nov/23

Look at the first few summands: i=0 → 1 i=1 → (n/(2n+4)) i=2 → ((n^2 −n)/(6n+24n+24)) i=3 → ((n^3 −3n^2 +2n)/(24n^3 +144n^2 +288n+192)) i=k → (n^k /((k+1)!n^k +...)) The limit_(n→∞) of the summands for i=k is (1/((k+1)!)) The sum then is Σ_(k=0) ^∞ (1/((k+1)!)) =e−1

$$\mathrm{Look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few}\:\mathrm{summands}: \\ $$$${i}=\mathrm{0}\:\rightarrow\:\mathrm{1} \\ $$$${i}=\mathrm{1}\:\rightarrow\:\frac{{n}}{\mathrm{2}{n}+\mathrm{4}} \\ $$$${i}=\mathrm{2}\:\rightarrow\:\frac{{n}^{\mathrm{2}} −{n}}{\mathrm{6}{n}+\mathrm{24}{n}+\mathrm{24}} \\ $$$${i}=\mathrm{3}\:\rightarrow\:\frac{{n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}}{\mathrm{24}{n}^{\mathrm{3}} +\mathrm{144}{n}^{\mathrm{2}} +\mathrm{288}{n}+\mathrm{192}} \\ $$$${i}={k}\:\rightarrow\:\frac{{n}^{{k}} }{\left({k}+\mathrm{1}\right)!{n}^{{k}} +…} \\ $$$$\mathrm{The}\:\underset{{n}\rightarrow\infty} {\mathrm{limit}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{summands}\:\mathrm{for}\:{i}={k}\:\mathrm{is}\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!} \\ $$$$\mathrm{The}\:\mathrm{sum}\:\mathrm{then}\:\mathrm{is}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\:=\mathrm{e}−\mathrm{1} \\ $$

Commented by MrGHK last updated on 29/Nov/23

$${very}\:{great} \\ $$

Answered by witcher3 last updated on 29/Nov/23

Σ_(i=0) ^n ((1/(n+2)))^i .((n!)/((n−i)!.i!(i+1))) Σ_(i=0) ^n .( ((n),(i) )/((n+2)^i (i+1))) (1+x)^n =Σ_(i=0) ^n ((n),(i) ).x^i ⇒∫_0 ^t (1+x)^n dx=Σ_(i=0) ^n ( ((n),(i) )/(i+1))t^(i+1) =(((1+t)^(n+1) −1)/(n+1)) ⇒Σ_(i=0) ^n ( ((n),(i) )/(i+1))((1/(n+2)))^i =(((1+(1/(n+2)))^(n+1) −1)/((n+1)))(n+2) =(e^((n+1)((1/(n+2))−(1/(2(n+2)^2 ))+o((1/n^2 )))) /1).((n+2)/(n+1));((n+2)/(n+1))∼1 ∼(e^(1−(1/(2(n+2)))+o((1/(n )))) −1)∼e−1 lim_(n→∞) .((n!)/((n−i)!.(i+1).(n+2)^i ))=e−1

$$\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)^{\mathrm{i}} .\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{i}\right)!.\mathrm{i}!\left(\mathrm{i}+\mathrm{1}\right)} \\ $$$$\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}.\frac{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{i}}\end{pmatrix}}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{i}} \left(\mathrm{i}+\mathrm{1}\right)} \\ $$$$\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} =\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{i}}\end{pmatrix}.\mathrm{x}^{\mathrm{i}} \Rightarrow\int_{\mathrm{0}} ^{{t}} \left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} \mathrm{dx}=\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{i}}\end{pmatrix}}{\mathrm{i}+\mathrm{1}}{t}^{{i}+\mathrm{1}} =\frac{\left(\mathrm{1}+{t}\right)^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\Rightarrow\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\begin{pmatrix}{\mathrm{n}}\\{\mathrm{i}}\end{pmatrix}}{\mathrm{i}+\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)^{\mathrm{i}} =\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} −\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)}\left(\mathrm{n}+\mathrm{2}\right) \\ $$$$=\frac{\mathrm{e}^{\left(\mathrm{n}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\right)} }{\mathrm{1}}.\frac{\mathrm{n}+\mathrm{2}}{{n}+\mathrm{1}};\frac{\mathrm{n}+\mathrm{2}}{\mathrm{n}+\mathrm{1}}\sim\mathrm{1} \\ $$$$\sim\left({e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{2}\right)}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}\:}\right)} −\mathrm{1}\right)\sim\mathrm{e}−\mathrm{1} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{n}!}{\left(\mathrm{n}−\mathrm{i}\right)!.\left(\mathrm{i}+\mathrm{1}\right).\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{i}} }=\mathrm{e}−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Commented by MrGHK last updated on 29/Nov/23

$${very}\:{nice}\:{solution} \\ $$

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