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Question Number 192276 by York12 last updated on 13/May/23

$$\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{k=0}{\sum ^n}\left[\frac{k(n-k)!+(k+1)}{(k+1)!(n-k)!}\right]\right)$$

Answered by witcher3 last updated on 14/May/23

$$\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}!(\mathrm{n}-\mathrm{k})!}=\frac{1}{\mathrm{n}!}\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{\mathrm{n}!}{\mathrm{k}!(\mathrm{n}-\mathrm{k})!}$$$$=\frac{1}{\mathrm{n}!}.\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}=\frac{2^{\mathrm{n}}}{\mathrm{n}!}$$$$\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\left(\frac{\mathrm{k}(\mathrm{n}-\mathbf{k})!+(\mathrm{k}+1)}{(\mathrm{k}+1)!(\mathrm{n}-\mathrm{k})!}\right)=\mathrm{\Sigma }\frac{1}{\mathrm{k}!}+\frac{1}{\mathrm{k}!(\mathrm{n}-\mathrm{k})!}$$$$=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}!}+\frac{2^{\mathrm{n}}}{\mathrm{n}!}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }=\mathrm{e}$$$$$$

Commented by York12 last updated on 20/Aug/23

$$$$$$\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{k=1}{\sum ^n}\left[\frac{k(n-k)!+(k+1)}{(k+1)!(n-k)!}\right]\right)=\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{k=1}{\sum ^n}[\frac{k}{(k+1)!}+\frac{1}{\mathrm{k}!(n-k)!}]\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }\left(\underset{k=1}{\sum ^n}[\frac{1}{(k!)}-\frac{1}{(k+1)!}]\right)+\underset{n\to \mathrm{\infty }}{\lim }\left(\frac{2^n}{n!}\right)=\underset{n\to 0}{\lim }(1-\frac{1}{(n+1)!}+0)=1\to \left({That}^{\prime }sit\right)$$$$$$

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