Find sum of series 2

Hong Quy · Yang-Mills Theory Offline Joined: 20 Feb 2005 Posts: 543

Find value of the sum:
$\sum_{n=0}^{ \infty} \frac{ (-1)^n(n+1)}{(n+2) n! 2^{n-1}}$

**Posted:** Sat Oct 31, 2009 3:14 am

_________________
wto_tientethegioi@yahoo.com
mathvn.org

Kent Merryfield

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

Integrate that, getting $g(x)$ such that $g'(x) = f(x)$ and

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

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

Integrate that, getting $h(x)$ such that $h'(x) = \frac {g(x)}x.$

$h(x) = \sum_{n = 0}^{\infty}\frac {( - 1)^nx^{n + 1}}{(n + 2)!} = \sum_{n = 2}^{\infty}\frac {( - 1)^nx^{n - 1}}{n!} = \frac ...$

$h'(x) = \frac {g(x)}x = - \frac {e^{ - x}}x + \frac {1 - e^{ - x}}{x^2}$

$g(x) = - e^{ - x} + \frac {1 - e^{ - x}}{x}$

$f(x) = g'(x) = e^{ - x} + \frac {xe^{ - x} -1+e^{ - x}}{x^2} = e^{ - x}\left(1 + \frac1x + \frac1{x^2}\right) - \frac1{x^2}.$

Our limit is $2f\left(\frac12\right) = \frac{14}{\sqrt {e}} -8.$

Stank

I do not understand your solution. However, I checked on two computer algebra systems. The first (TI-nspire CAS) returned $\displaystyle\sum_{n=0}^{\infty}{\frac{(n+1)\cos{n\pi}{2}^{-n}}{(n+2)n!}$ . The other (Wolfram Mathematica) returned $\frac{14-9\sqrt{e}}{\sqrt{e}}$ . This may help.

**Posted:** Sat Oct 31, 2009 12:17 pm

_________________
Check out or post on my computer generated Math/Science image gallery! http://www.artofproblemsolving.com/Forum/album.php?t=306615

"Your solution may be faster, but the quadratic formula's for squares!" -SP

Kent Merryfield

$\frac{14-9\sqrt{e}}{\sqrt{e}}$ can't be right, since it's negative. This is an alternating series and its sum is positive, and numerically equal to about $0.49142924.$

My own answer is also wrong. I can see where I'm off by a factor of 2, since the answer should be $2f\left(\frac12).$ But that only makes my mistake worse.

The method should work; there's a mistake in the details somewhere. I'm still looking.

**Posted:** Sat Oct 31, 2009 12:57 pm

JBL

An alternative approach, since Kent has already claimed power series methods:
$\begin{align*} \sum_{n = 0}^{\infty}\frac {( - 1)^{n}(n + 1)}{(n + 2)n!2^{n - 1}} & = \sum_{n = 0}^{\infty}\frac {( - 1)^...$

Stank, you probably started your sum at $n = 1$ instead of $n = 0$ .
Kent, it looks like something went wrong when you took the derivative of $g(x)$ .