# Thread: Find the sum of the series

1. ## Find the sum of the series

Find the sum of the series,

$\displaystyle \sum^{\infty}_{n=1} \left( \frac{n+2}{n!} \right)x^{n+1}$

See figure attached for my attempt.

EDIT: I tried taking another route in my work, see 2nd figure attached for that attempt. (Follow the arrow I added in)

2. Originally Posted by jegues
Find the sum of the series,

$\displaystyle \sum^{\infty}_{n=1} \left( \frac{n+2}{n!} \right)x^{n+1}$

See figure attached for my attempt.
Are you sure you want the sum and not something else (like the interval of convergence)?

3. Originally Posted by mr fantastic
Are you sure you want the sum and not something else (like the interval of convergence)?
The only thing the question states is Find the sum of the series. I've added another attempt in the 2nd figure of my OP.

4. Originally Posted by jegues
Find the sum of the series,

$\displaystyle \sum^{\infty}_{n=1} \left( \frac{n+2}{n!} \right)x^{n+1}$

See figure attached for my attempt.

EDIT: I tried taking another route in my work, see 2nd figure attached for that attempt. (Follow the arrow I added in)
Let consider the series...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{x^{n+2}}{n!} = x^{2}\ \sum_{n=1}^{\infty} \frac{x^{n}}{n!} = x^{2}\ (e^{x}-1)= f(x)$ (1)

Very well!... from (1) You obtain directly...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{(n+2)}{n!}\ x^{n+1} = f^{'} (x) = x\ (x+2)\ e^{x} -2 x$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Originally Posted by chisigma
Let consider the series...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{x^{n+2}}{n!} = x^{2}\ \sum_{n=1}^{\infty} \frac{x^{n}}{n!} = x^{2}\ (e^{x}-1)= f(x)$ (1)

Very well!... from (1) You obtain directly...

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{(n+2)}{n!}\ x^{n+1} = f^{'} (x) = x\ (x+2)\ e^{x} -2 x$ (2)
Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Aha! So my 2nd attempt was correct!

Thanks again!