-1

I need to prove that $x_n = \frac{3^n}{n!}$ converges and to find the limit. I am wondering if I can use the squeeze theorem for this problem. If so, how would I structure my proof?

Julia
  • 29
  • https://math.stackexchange.com/questions/1828346/prove-lim-n-to-infty-fracann-0 – zkutch Nov 02 '20 at 00:46
  • Hi Julia!, welcome to MSE. Dear @zkutch, I believe OP wanted her question to be answered using limit definitions. I do not understand though, why there is a tag for complex-analysis. –  Nov 02 '20 at 00:49
  • Hi Julia! Please read How to ask a good question on Math.se ASAP – amWhy Nov 02 '20 at 00:56
  • Dear Julia, its easy to show that $x_{n}$ is increasing now all you need to do is show that limit of $\frac{3^{n}}{n!}$ is bounded, to do so try to use a similar technique to showing that $\lim_{n\to+\infty} \frac{1}{n!}$ is bounded as such $2<e<3$. It will work for $\frac{3^{n}}{n!}$. Finally, I would hope that users would not close this question because OP is new and deserves to be assisted. –  Nov 02 '20 at 01:07
  • @T.H.Shehadi your comment is worthless because you seem to have conflated series and sequences. – Integrand Nov 02 '20 at 01:40
  • @Integrand The OP wrote in title series but in the question the OP wrote a sequence! –  Nov 02 '20 at 01:41
  • I voted to re-open, despite the fact that the query has been definitely answered on this webpage, because I dispute that the query is a duplicate of the query cited. The reason for my disagreement (and I could be mistaken) is that the OP indicated "...and to find the limit." Perhaps I overlooked something, but in the cited query, I did not notice any explicit calculation of the actual limit that the series converges to. – user2661923 Nov 02 '20 at 02:59

1 Answers1

0

Let $ e^x = \sum_{n=0}^{\infty} a_n x^n $, Sinc \frac{d}{dx} (e^x) = e^x, then

$$\frac{d}{dx} (\sum_{n=0}^{\infty} a_n x^n ) = \sum_{n=0}^{\infty} a_n x^n .$$

And

$$\sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} a_n x^n, $$

equaling the coefficient of the x that has the same power, we get

$$a_{n} = \frac{1}{n} = a_{n-1},$$

so,

$$a_n = \frac{1}{n} a_{n-1} = \frac{1}{n} \frac{1}{n-1} a_{n-2}=....= \frac{1}{n!}a_0.$$

Since $e^0 = 1$ then we have $a_0 =1$. Thus $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, you need just to replace $x$ by 3.

Mr. Proof
  • 1,702