# Nasty Series: sum n/(n+1)!

1. Jul 29, 2008

### 3029298

The problem

Calculate the following sum:

$$\sum_{n=1}^{\infty}\frac{n}{\left(n+1\right)!}$$

The solution

I know the solution of the partial sum:

$$\sum_{n=1}^{K}\frac{n}{\left(n+1\right)!}=\frac{\left(K+1\right)!-1}{\left(K+1\right)!}$$

If we take the limit of $$K\rightarrow\infty$$ the sum is equal to 1.

This solution can then be easily proved by induction.

But my question is: how can you obtain a solution without knowing this answer and proving it by induction?

Any help would be really appreciated!

2. Jul 29, 2008

### Dick

Let f(x)=exp(x)/x and consider the derivative of the taylor series of f(x) evaluated at x=1. It's -1+S where S is your series. Now directly evaluate f'(1). What do you conclude S is??

3. Jul 29, 2008

### 3029298

The derivative of the Taylor series you mention, looks like this:

$$e\left(\left(x-1\right)-\left(x-1\right)^2+\frac{3}{2}\left(x-1\right)^3-\frac{11}{6}\left(x-1\right)^4+....\right)$$

I do not see anything emerging from this... :shy:

4. Jul 29, 2008

### Dick

The series expanded around x=0. I.e. f(x)=(1/x)*(1+x+x^2/2!+x^3/3!+...). Multiply the 1/x in and differentiate and put x=1.

5. Jul 29, 2008

### 3029298

But then the term in the numerator remains 1. In the series it is n, so this does not lead to the correct answer. We in fact want the sum of:

$$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+.....+\frac{n}{(n+1)!}$$

6. Jul 29, 2008

### Dick

f(x)=1/x+1+x/2!+x^2/3!+x^3/4!+... f'(x)=-1/x^2+1/2!+2x/3!+3x^2/4!+... Put x=1. Do you see your series now?? The n in the numerator comes from the differentiation.

7. Jul 29, 2008

### 3029298

wow, this is very cool!

now I'm interested in how you actually thought of this solution!!
you could really help me by showing what strategy you use to solve the series with factorials in them, because those are the ones I have the most difficulties with.

8. Jul 29, 2008

### D H

Staff Emeritus
Denote the series as S. Since the terms are all positive, the series is absolutely convergent (if it converges). Add e-2[/tex] to S and rearrange terms (valid iff S converges): \begin{aligned} S + e-2 &= S + \sum_{n=2}^{\infty}\frac 1 {n!} \\ &= S + \sum_{n=1}^{\infty}\frac 1 {(n+1)!} \\ &= \sum_{n=1}^{\infty}\frac{n+1}{(n+1)!} \\ &= \sum_{n=1}^{\infty}\frac{1}{n!} \\ &= e-1 \end{aligned} Thus [itex]S+e-2=e-1[/tex], or $$S=1. 9. Jul 29, 2008 ### 3029298 beautiful! thanks a lot! 10. Jul 29, 2008 ### Dick I thought of it by considering the general term in the expansion of exp(x). x^n/n!. I can get an n in the top by differentiating. But then I get n/n!. I want (n-1)/n!. So I divided by x first. That's all. It's not a very general technique, but if you can recognize a series as 'looking like' a power series, it works. 11. Jul 29, 2008 ### 3029298 thanks for all this help! I learned a lot today 12. Jul 29, 2008 ### D H Staff Emeritus Finding a general expression for a partial sum by induction and then finding the limit of this partial sum is a perfectly valid technique. Dick and I both used tricks. The partial sum approach of course involves a "trick" as well -- finding an expression for the dang partial sum. 13. Jul 29, 2008 ### 3029298 yeah exactly, and since I found the partial sum somewhere on the internet, I prefer a healthier approach :p 14. Jul 29, 2008 ### 3029298 Oh, this is GREAT! By using this trick I could now do the following exercises as well: [tex] \sum_{n=0}^\infty\frac{1}{(2n)!}$$

$$\sum_{n=0}^\infty\frac{1}{(2n+1)!}$$

$$\sum_{n=1}^\infty\frac{n}{(2n+1)!}$$

by considering the sinus hyperbolicus and cosinus hyperbolicus! Thanks a lot for this big eye-opener!!

15. Jul 29, 2008

### Dick

Atta boy! Go nuts!