0
$\begingroup$

From what value of $a$ will the following series be divergent: $$\sum_{n=0}^{\infty} \dfrac{a^{n}}{n!}$$

  • 0
    For none: that thing's always convergent for values in $\;\Bbb C\;$ . – 2017-01-15
  • 0
    Does $C$ represent set of real numbers or complex numbers? – 2017-01-15
  • 0
    @S Yup...but if you only want the reals it is just the very same argument mutatis mutandis. – 2017-01-15
  • 0
    I get it. Even though I was not after complex nums – 2017-01-15

4 Answers 4

1

This series is convergent for all $a \in \mathbb{R}$ and your sum is $e^a$, because $e^x=\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ for all $x \in \mathbb{R}$.

4

Try D'Alembert's Ratio Test:

$$\lim_\limits{n\to\infty}\left|\frac{x_{n+1}}{x_n}\right|=\lim_\limits{n\to\infty}\left|\frac{\dfrac{a^{n+1}}{(n+1)!}}{\dfrac{a^{n}}{n!}}\right|$$ $$=\lim_\limits{n\to\infty}\left|\frac{a}{n+1}\right|=0<1$$

The series converges for all $a \in \mathbb{R}$.

  • 0
    @DonAntonio Ah yes! Thanks for pointing out. – 2017-01-15
2

Clearly the series converges for $\;a=0\;$ , and for $\;a\neq0\;$ we can use the $\;n\,-$th root test:

$$\lim_{n\to\infty}\sup\sqrt[n]{\frac{|a|^n}{n!}}=0\implies R=\infty$$

with $\;R=$ the convergence radius, so the series converges for all $\;a\in\Bbb C\;$ .

In fact, this is the series for $\;e^a\;$ ...

1

Let $$u_n=\frac{a^n}{n!}.$$ Then $$\left|\frac{u_{n+1}}{u_n} \right|=\left|\frac{a^na}{(n+1)n!}\cdot\frac{n!}{a^n}\right|=\frac{1}{n+1}|a|\to 0 \text{ as } n\to\infty$$

So, using the Ratio Test, the given series converges for all values of $a$. Therefore, no such value of $a$ the given series diverges.