For what values of $x$ is the $\sum_{n=1}^\infty \frac{(7x)^n}{n!}$ convergent?

Question

I was wondering for which values of $x$ the following series converges: $$\sum_{n=1}^\infty \frac{(7x)^n}{n!}.$$

I applied the ratio test to get $$\lim_{n\to \infty} \frac{7x}{n+1}$$

So then $7|x|<1$ so $|x|<\frac{1}{7}$ but apparently this is wrong? Can someone see where I have gone wrong?

Thank you!

Answer 1

It is wrong because that limit is always $0$, and not only when $7\lvert x\rvert<1$. Therefore, the series always converges.

Answer 2

Recall that:

$$\sum_{n=0}^\infty \frac{x^n}{n!} = e^x,$$

which is known to be convergent for all $x$.

Therefore, your series converges and its value is equal to:

$$\sum_{n=1}^\infty \frac{(7x)^n}{n!} = \sum_{n=0}^\infty \frac{(7x)^n}{n!} - \frac{(7x)^0}{0!} =e^{7x} - 1.$$

Answer 3

I applied the ratio test to get $$\lim_{n\to \infty} \frac{7x}{n+1}$$

Consider $x$ fixed, what is this limit for $n \to \infty$? Hence the series converges for ...