# Thread: sum of a series

1. ## sum of a series

It's been awhile since I have done these types of problems.

Can the sum of a series have a factorial in the answer?

this is my problem: Summation of (n/(n+1)!) with n>= 1

while working it out, I found that this is 1/(n+1)(n-1)!

2. Originally Posted by Nichelle14
It's been awhile since I have done these types of problems.

Can the sum of a series have a factorial in the answer?

this is my problem: Summation of (n/(n+1)!) with n>= 1

while working it out, I found that this is 1/(n+1)(n-1)!

Are you familiar with the exponential series, expansion of $e^x$?

KeepSmiling
Malay

3. Consider a function,
$
f(x)=\frac{x^2}{2!}+\frac{2x^3}{3!}+\frac{3x^4}{4! }+\frac{4x^5}{5!}+...
$

Its derivative is, (since it is absolutely convergent)
$f'(x)=x+\frac{x^2}{1}+\frac{x^3}{2!}+\frac{x^4}{3! }+...$
Thus, (since it is abolsulte convegent)
$f'(x)=x\left( 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+... \right)$
Now, (as malaygoel said) are you familiar with,
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
Then you are left with,
$f'(x)=xe^x$
Thus, to solve this diffrencial equation find,
$\int xe^x dx$ use integration by parts with,
$u=x$ then, $u'=1$ and $v'=e^x$ thus, $v=e^x$.
Thus, (by parts)
$xe^x-\int e^xdx =xe^x-e^x+C$
Thus,
$f(x)=xe^x-e^x+C$
we see that, $f(0)=0$
Thus,
$0e^0-e^0+C=0$ thus, $C=1$
Finally our required function is,
$f(x)=xe^x-e^x+1$
Note, the infinite sum, as you said,
$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...$
Is, the same as $f(1)$ which in this case is,
$1e^1-e^1+1=1$

4. I vaguely remember the exponential series of e^x.

This may be a silly question but what exactly is the answer? Do I need to do more?

Also, when you were trying to find C, why did you set the function equal to 1?

5. Originally Posted by Nichelle14
I vaguely remember the exponential series of e^x.

2, because that series as I explained is $f(1)=1$.