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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!

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  • $\begingroup$ Replace $7x$ by $X$. What about the convergence of $\exp(X)$? $\endgroup$ – Dietrich Burde Apr 24 '19 at 14:16
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    $\begingroup$ $\lim_{n\to\infty}\frac{7x}{n+1}=0$ $\endgroup$ – kingW3 Apr 24 '19 at 14:17
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It is wrong because that limit is always $0$, and not only when $7\lvert x\rvert<1$. Therefore, the series always converges.

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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:

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

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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 ...

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