QUESTION: Find the value of- $$\lim_{x \rightarrow -\infty}\sum_{k=1}^{1000} \frac{x^k}{k!}$$
MY ANSWER: Since $x→-\infty$ therefore the summation will look like-
$$ \frac{-\infty^1}{1!}+ \frac{-\infty^2}{2!} + \frac{-\infty^3}{3!} + \frac{-\infty^4}{4!}$$
Now, we know that $${-\infty^{even}}=\infty$$
And $1000$ can be broken down into 500 pairs. And since $k$ is finite the terms will look like $$ -\infty + \infty -\infty +\infty-…..$$
Hence, we can conclude that the overall limit of the summation will tend to zero.
Am I correct?
Thank you for your help.
Best Answer
One may observe that, for $x\neq0$, $$ \sum_{k=1}^{1000} \frac{x^k}{k!}=\color{red}{\frac{x^{1000}}{1000!}}\times\left(\color{red}1+\frac{1000!}{x\times999!}+\frac{1000!}{x^2\times998!}+\cdots+ \frac{1000!}{x^{1000}\times1!}\right) $$ giving that $$ \begin{align} \lim_{x \rightarrow -\infty}{\sum_{k=1}^{1000} \frac{x^k}{k!}} &=\lim_{x \rightarrow -\infty} \color{red}{\frac{x^{1000}}{1000!}}\times \lim_{x \rightarrow -\infty}\left(\color{red}1+\frac{1000!}{x\times999!}+\frac{1000!}{x^2\times998!}+\cdots+ \frac{1000!}{x^{1000}\times1!}\right) \\\\&=\lim_{x \rightarrow -\infty} \color{red}{\frac{x^{1000}}{1000!}}\times \color{red}1 \\\\&=\infty. \end{align} $$