# power series of function

I am studying on summation theory on power series of functions. My question is to find the sum of power series $$e^{-n} \sum_{k=0}^{\infty} \frac{n^k}{k!}\frac{k}{k+1}.$$ I tried apply differentiation and integration but could not find anything useful. Thanks in advance from now.

• meta.math.stackexchange.com/questions/5020/… There is a very weird-looking thing in your question: (n^k).(k/k+1) ... Is this a fraction or what? Better to write this with MathJax. – Timbuc Dec 17 '14 at 20:19
• consider this way e^(-n)Σ((n^k)(x_k))/k! where (x_k)=(k/k+1) i am sorry i am new to this website and math i will learn mathtyping asap. I hope things are clear now. – leo Dec 17 '14 at 20:26
• you are missing k in the numerator. could you please add it...thanks for helping write it nicely – leo Dec 17 '14 at 20:29
• I meant you miss k when you wrote the series more managable should not it be (n^k)k/(k+1)! – leo Dec 17 '14 at 20:40

You are starting with $$e^{-x} \sum_{k = 1}^\infty \frac{kx^k}{(k+1)!} = e^{-x}F(x),$$ which is almost very easy. Let's ignore the $e^{-x}$ piece, because it's not interesting. So we just consider $F(x)$.
Notice that $\dfrac{F(x)}{x} = \displaystyle\sum_{k = 1}^\infty \frac{kx^{k-1}}{(k+1)!}.$ If we integrate this, we see that \begin{align} \int_0^x \frac{F(t)}{t} dt &= C + \sum_{k = 1}^\infty \frac{x^k}{(k+1)!} = C + \frac{1}{x}\sum_{k = 2}^\infty \frac{x^k}{k!} \\ &= C - \frac{1}{x} - 1 + \frac{1}{x} \sum_{k = 0}^\infty \frac{x^k}{k!} \\ &= \frac{e^x}{x} - \frac{1}{x} - 1 + C. \end{align}
We are interested in $F(x)$, which we recover from the fundamental theorem of calculus. \begin{align} \frac{d}{dx} \int_0^x \frac{F(t)}{t} dt &= \frac{F(x)}{x} \\ &= \frac{d}{dx} \left( \frac{e^x}{x} - \frac{1}{x} - 1 + C \right) \\ &= \frac{e^x}{x} - \frac{e^x}{x^2} + \frac{1}{x^2} \end{align}
Thus $$F(x) = e^x - \frac{e^x}{x} + \frac{1}{x},$$ so that $$e^{-x}F(x) = 1 - \frac{1}{x} + \frac{e^{-x}}{x},$$ which is your answer. $\diamondsuit$