Finding the Limit of a Sum with Factorials

[namu]

Homework Statement

Find the limit

$lim_{n \to \infty} \sum_{j=1}^n \frac{b^j}{(j+1)!}$

Homework Equations

Geometric series sum:

$S=\sum_{j=1}^n r^n$

$S-rS=(1-r)S=1-r^{n+1}$

$S=\frac{1-r^{n+1}}{1-r}$

$S \to \frac{1}{1-r} \,\,\, as \,\,\, n \to \infty$

if $|r|<1$

The Attempt at a Solution

$b\sum_{j=1}^n \frac{b^j}{(j+1)!}-\sum_{j=1}^n \frac{b^j}{(j+1)!}=-\frac{b}{2}+\frac{b^2}{3}+\frac{b^3}{8}+...$

I tried to use something similar as when deriving the sum of a geometric series, however was unsucessful. I don't know how to integrate a factorial, so I can't use that approach either. Does anyone have any suggestions?

 

[Dick]

You can forget about the geometric series. That's no help at all. Suppose the problem were $lim_{n \to \infty} \sum_{j=1}^n \frac{b^j}{(j)!}$. Could you do that one?

 

[Deveno]

is it possible that this might represent the taylor series for some function f at b?

if so, what function do you think it might be?

 

[namu]

hmmm...it might just be a Taylor series. If this was

$$\sum_{n=0}^{\infty} \frac{b^j}{j!}$$

then this will be the function

$$exp(b)$$

The sum runs from n=1, so something like

$$exp(b)-1$$

still not quite sure what function would result in such a series.

 

[SammyS]

hmmm...it might just be a Taylor series. If this was
$$\sum_{n=0}^{\infty} \frac{b^j}{j!}$$
then this will be the function
$$exp(b)$$
The sum runs from n=1, so something like
$$exp(b)-1$$
still not quite sure what function would result in such a series.

That's a start.

Keep working with your sum like you have been.

Here's your sum:

$\displaystyle \sum_{j=1}^n \frac{b^j}{(j+1)!}=\frac{b}{2!}+\frac{b^2}{3!}+ \frac{b^3}{4!}+\frac{b^4}{5!}+\dots$​

Here's what you have for e^b:

$\displaystyle \sum_{n=0}^{\infty} \frac{b^j}{j!}=1+\frac{b}{1!}+\frac{b^2}{2!}+ \frac{b^3}{3!}+\frac{b^4}{4!}+\dots$​

What if you divide e^b by b?

 

[namu]

Yes, that is exactly it. It is

$$\frac{e^b-1-b}{b}$$

Thank you.

 

[SammyS]

Yes, that is exactly it. It is

$$\frac{e^b-1}{b}$$

Thank you.

Recheck your result. I think it's not quite right.

 

[namu]

Yes, silly mistake. I fixed it. See above. Thank you.