While studying for the Michigan Mathematics Prize Competition, I came across this problem on an older exam:

 

The value of [math] \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{2006}{2007!} [/math] is

 

A) [math]0.9999[/math]

B) [math]1[/math]

C) [math]\frac{2006}{2007}[/math]

D) [math]1-\frac{1}{2007!}[/math]

E) [math]\frac{1}{2007!}[/math]

 

When I went to evaluate the series I did the following:

[math] \frac{1}{2}+\frac{1}{3}+\frac{1}{8}....[/math]

So I know it cannot be found using the summation for a geometric or arithmetic sequence. However I was easily able to eliminate answer E. I was however able to write the sequence in summation notation:

[math]\sum \frac{i}{(i+1)!}[/math]

This didn't help me solve the problem. Can anyone please show me how to do this

The answer is D)

 

Proof

 

Lets do it slightly more generally. Let us consider

 

[math]\sum_{i=1}^{n} \frac{i}{(i+1)!}[/math].

 

Then let us use the answer D). Thus we want to show that

 

[math]\sum_{i=1}^{n} \frac{i}{(i+1)!} + \frac{1}{(n+1)!} = 1[/math].

 

By expanding out term by term in of the first part of the left hand side, using the definition of the factorial you can see that

 

[math] \sum_{i=1}^{n} \frac{i}{(i+1)!} + \frac{1}{(n+1)!} = \sum_{i=1}^{n-k}\frac{i}{(i+1)!} + \frac{1}{(n-k +1)!}[/math],

 

for any suitable [math]k[/math].

 

Then put in [math]k = n-1[/math] into the above and you get

 

[math]\frac{1}{2!} + \frac{1}{(n-n+1 +1)} =\frac{1}{2!} + \frac{1}{2!} = 1 [/math].

 

Thus we have

 

[math]\sum_{i=1}^{n} \frac{i}{(i+1)!} = 1- \frac{1}{(n+1)!}[/math].

 

 

Hope that helps. You can "flesh it out yourself".