How can i solve the series #\sum_{n=0}^{\infty} \frac{n^3}{(n+1)!} # ?

1 Answer
Jul 27, 2018

Calling #f(x) = \frac{e^x-1}{x} = \sum_{k=0}^{\infty}\frac{x^k}{(k+1)!}# we have

# x\frac{d}{dx}\(x\frac{d}{dx}(x\frac{d}{dx}f(x))\) = \sum_{k=0}^{\infty}k^3 \frac{x^k}{(k+1)!} = \frac{e^x \(x^3+x-1\)+1}{x} #

now making #x=1# we have

# \sum_{k=0}^{\infty} \frac{k^3}{(k+1)!} = e+1 #