Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

Commented by Frix last updated on 29/Nov/23

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

Commented by MrGHK last updated on 29/Nov/23

$$verygreat$$

Answered by witcher3 last updated on 29/Nov/23

$$\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}}}{\left(\frac{1}{\mathrm{n}+2}\right)}^{\mathrm{i}}.\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{i})!.\mathrm{i}!(\mathrm{i}+1)}$$$$\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}}}.\frac{\left(\begin{array}{c}\mathrm{n}\\ \mathrm{i}\end{array}\right)}{{(\mathrm{n}+2)}^{\mathrm{i}}(\mathrm{i}+1)}$$$${(1+\mathrm{x})}^{\mathrm{n}}=\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{i}\end{array}\right).{\mathrm{x}}^{\mathrm{i}}\Rightarrow {\int }_0^t{(1+\mathrm{x})}^{\mathrm{n}}\mathrm{dx}=\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}}}\frac{\left(\begin{array}{c}\mathrm{n}\\ \mathrm{i}\end{array}\right)}{\mathrm{i}+1}{t}^{i+1}=\frac{{(1+t)}^{n+1}-1}{n+1}$$$$\Rightarrow \underset{\mathrm{i}=0}{\sum ^{\mathrm{n}}}\frac{\left(\begin{array}{c}\mathrm{n}\\ \mathrm{i}\end{array}\right)}{\mathrm{i}+1}{\left(\frac{1}{\mathrm{n}+2}\right)}^{\mathrm{i}}=\frac{{(1+\frac{1}{\mathrm{n}+2})}^{\mathrm{n}+1}-1}{(\mathrm{n}+1)}(\mathrm{n}+2)$$$$=\frac{{\mathrm{e}}^{(\mathrm{n}+1)(\frac{1}{\mathrm{n}+2}-\frac{1}{2{(\mathrm{n}+2)}^2}+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right))}}{1}.\frac{\mathrm{n}+2}{n+1};\frac{\mathrm{n}+2}{\mathrm{n}+1}\sim 1$$$$\sim (e^{1-\frac{1}{2(\mathrm{n}+2)}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)}-1)\sim \mathrm{e}-1$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }.\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{i})!.(\mathrm{i}+1).{(\mathrm{n}+2)}^{\mathrm{i}}}=\mathrm{e}-1$$$$$$$$$$

Commented by MrGHK last updated on 29/Nov/23

$$verynicesolution$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com