Convergence of $\sum_{n =0}^{+\infty} \frac{3^n}{n!}$ with geometric progression

Question

So, I have

$$\sum_{n = 0}^{+\infty} \dfrac{3^n}{n!}$$

I know that the series converges, and I know the sum is $e^3$. Yet I wanted to prove this with the factorial majorization and the user of the geometric progression. I'm however stuck, I tried basically anything I could.

I started with considering

$$\sum_{n = 0}^{+\infty} s_n = \lim_{k\to +\infty} s_k = \lim_{k\to +\infty} \sum_{n = 0}^k s_k$$

Now:

$$s_n = 1 + 3 + \frac{3^2}{2!} + \frac{3^3}{3!} + \cdots + \frac{3^{n-1}}{(n-1)!} + \frac{3^n}{n!}$$

I wanted to use the fact that (proof by induction: made) for $n\geq 3$ we have

$$n! \geq 3^{n-2}$$ and then $$\frac{1}{n!} \leq \frac{1}{3^{n-2}}$$

But from here I have no idea of how to obtain a geometric progression both to prove the convergence and to give a rough estimation of the value of the sum. I always end up in some nonsense.

Also because of the numerator. If I had just $\frac{1}{n!}$ then I could do something like:

$$1 + \frac{1}{2} + \cdot \frac{1}{n!}$$ using $n! \geq 2^{n-1}$ for $n\geq 2$ that is

$$1 + \frac{1}{2} + \frac{1}{3!} + \ldots \leq 1 + \frac{1}{2} + \frac{1}{2^2} + \cdot + \frac{1}{2^{n-1}} = \sum_{k=0}^{n} \frac{1}{2^k} - 1$$ and then use the geometric progression.

But with that $3^n$ and so on at the numerator, I'm stuck.

How to proceed?

Answer 1

Since $n! \geq 6 * 4^{n-3}$ for all $n \geq 4$ you can bound the tail of this series geometrically: $$\sum_{k=4}^\infty \frac{3^k}{k!} \leq C \sum_{k=4}^\infty \frac{3^k}{4^k} < \infty,$$ where $C$ is a suitably chosen constant.

Answer 2

$$\begin{matrix}\sum\limits_{n = 0}^{+\infty} \dfrac{3^n}{n!} &=& \sum\limits_{n = 0}^{k-1} \dfrac{3^n}{n!}&+&\sum\limits_{n = k}^{+\infty} \dfrac{3^n}{n!}\\ & \le & \sum\limits_{n = 0}^{k-1} \dfrac{3^n}{n!}&+&\dfrac{3^k}{k!}\sum\limits_{n = k}^{+\infty} \left(\dfrac{3}{k}\right)^{n-k} \\ &=& \sum\limits_{n = 0}^{k-1} \dfrac{3^n}{n!}&+&\dfrac{3^k}{k!}\dfrac{k}{k-3}\end{matrix}$$ when $k>3$, giving an upper bound. You could use $\sum\limits_{n = 0}^{k-1} \dfrac{3^n}{n!}$ as a lower bound.

If you try these bounds for various values of $k$, you get closer to $e^3\approx 20.08554$ as $k$ increases. The upper bound is closer to the actual result that the lower bound is, since the lower bound's error is more than $\frac{3^k}{k!}$.

   k  lower    upper
   4 13.00000 26.50000
   5 16.37500 21.43750
   6 18.40000 20.42500
   7 19.41250 20.17188
   8 19.84643 20.10679
   9 20.00915 20.09051
  10 20.06339 20.08664
  11 20.07967 20.08577
  12 20.08410 20.08558
  13 20.08521 20.08555
  14 20.08547 20.08554
  15 20.08552 20.08554
  16 20.08553 20.08554
  17 20.08554 20.08554