Sum of series by comparing to a known Series expansion

Question

I can't seem to quite grasp Summation of Series any bit and would like to ask for your help.

The problem requires me to find the sum of series by using known series expansions. I have given the problem a try however I'm not quite sure if it's how it's meant to be done.

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The first problem is as follows: $$1 + 5 + \frac{5^2}{2!} + \frac{5^3}{3!} + ...$$

This seems to fall under the exponential expansion: $$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$

Does it simply mean that the sum of the series is simply $e^5$?

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Other question seems to be a combination of more than one known expansion sin and cos: $$1 - 3 + \frac{3^2}{2!} - \frac{3^3}{3!}+....$$

I'm honestly not sure how the last one ought to be worked

Answer 1

For the first: Yes, it is $e^5$.

For the second: The general term seems to be $(-1)^n\frac{3^n}{n!}$. What happens if you combine $(-1)^n3^n$ into one power?

Answer 2

Consider the expansion $$e^x = 1+x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

What do you get substituting $x=5$ and $x=-3$?

Answer 3

Yes, since $$ e^x=1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... ,\qquad x \in \mathbb{R}, \tag1 $$ then $$ e^5=1 + 5 + \frac{5^2}{2!} + \frac{5^3}{3!} + ... , $$ You may notice that, substituting $x \to -x$ in $(1)$, gives $$ e^{-x}=1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ... ,\qquad x \in \mathbb{R}, \tag2 $$ then $$ e^{-3}=1 - 3 + \frac{3^2}{2!} - \frac{3^3}{3!} + ... . $$ You may also write, in a compact form $$ e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!} $$ or $$ e^{-x}=\sum_{k=0}^{\infty}(-1)^k\frac{x^k}{k!} $$