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

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!

• Replace $7x$ by $X$. What about the convergence of $\exp(X)$? – Dietrich Burde Apr 24 '19 at 14:16
• $\lim_{n\to\infty}\frac{7x}{n+1}=0$ – kingW3 Apr 24 '19 at 14:17

## 3 Answers

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

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:

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

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 ...