1

Prove that the sequence $a_n = \sum_{k=0}^n \frac{1}{k!}$ is convergent.

I am using the theorem: All bounded monotone sequences converge.

So i need to prove it is bounded and monotone.

$a_1=1,a_2=1.5, a_3=1.667, a_4=1.708, a_5=1.717, a_6=1.7181, a_7=1.7182$

I can see that it is bounded below by 1 and increasing, but I'm not sure how to go about proving so a little tip in the right direction there would be great.

As far as finding where it is bounded above I think I need to take the limit using a geometric series? but I'm not sure how to do that with the "!" in the problem. Is the geometric series thing the right direction to go in?

I apologize if I'm asking too much, please do not solve the problem for me. I only want tips so I know how to go about it.

Elizabeth Austin Griffith
  • 45
  • What kind of series do you know to be convergence? For instance do you know that $\sum \frac{1}{k^2}$ convergence? what about $\sum \frac{1}{2^k}$? – Yanko Nov 11 '18 at 20:51
  • 2
    That it's monotone is pretty obvious, we add positive terms. For boundedness, yes, you usually bound it above by geometric series – Jakobian Nov 11 '18 at 20:51

4 Answers4

5

HINT

We can use that $k!\ge k^2$ for $k\ge 4$ and therefore $\frac1{k!}\le \frac1{k^2}$.

user
  • 154,566
5

Another way: For $k \geq 2$,

$$ \frac{1}{k!} \leq \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k} $$

so that, for $n \geq 2$,

$$ \sum_{k=0}^{n} \frac{1}{k!} \leq 2 + \sum_{k=2}^{n} \frac{1}{k(k-1)} = 3 - \frac{1}{n} \leq 3. $$

Sangchul Lee
  • 167,468
  • That's indeed related to $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$. Very nice direct proof. – user Nov 11 '18 at 20:59
4

Hint :

If you don't know the convergence of : $\sum_{k = 1}^n \frac{1}{n^2}$ then you can use the fact that that : $$2^k \leq k! \Leftrightarrow \frac{1}{2^k} \geq \frac{1}{k!}$$

and we have a geometric serie which converge.

Thinking
  • 1,310
0

$$ \\a_n=2+\sum_{m=2}^{n}\frac{1}{m!}\le2+\sum_{m=2}^{n}\frac{1}{m(m-1)}=2+\sum_{m=2}^{n}(\frac{1}{m-1}-\frac{1}{m})=3-\frac{1}{n} \\0<a_n\le 3-\frac{1}{n}<3 \\0<\lim_{n\to+\infty}a_n\le3 $$

Samvel Safaryan
  • 580