For $k 0, \sum_{x=0}^{\infty} \frac{k^x}{x !} \lim _{n \rightarrow \infty} \frac{n !}{(n-x) !}\left(1-\frac{k}{n}\right)^{n-x}\left(\frac{1}{n}\right)^x=$

Question

For $k > 0, \sum_{x=0}^{\infty} \frac{k^x}{x !} \lim _{n \rightarrow \infty} \frac{n !}{(n-x) !}\left(1-\frac{k}{n}\right)^{n-x}\left(\frac{1}{n}\right)^x=$

Aicharya · Accepted Answer

$\begin{aligned} & \text {For } k > 0, \sum_{x=0}^{\infty} \frac{k^x}{x !} \lim _{n \rightarrow \infty} \frac{n !}{(n-x) !}\left(1-\frac{k}{n}\right)^{n-x}\left(\frac{1}{n}\right)^x \ & =\lim _{n \rightarrow \infty} \sum_{x=0}^n \frac{n !}{x !(n-x) !}\left(1-\frac{k}{n}\right)^{n-x}\left(\frac{k}{n}\right)^x \ & =\lim _{n \rightarrow \infty} \sum_{x=0}^n{ }^n C_x\left(1-\frac{k}{n}\right)^{n-x}\left(\frac{k}{n}\right)^x \ & =\lim _{n \rightarrow \infty}\left(1-\frac{k}{n}+\frac{k}{n}\right)^n=\lim _{n \rightarrow \infty} 1^n=1 \end{aligned}$ Hence, option (4) is correct.

For \(k > 0, \sum_{x=0}^{\infty} \frac{k^x}{x !} \lim _{n \rightarrow \infty} \frac{n !}{(n-x)…

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