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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.

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  • $\begingroup$ 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. $\endgroup$ – Timbuc Dec 17 '14 at 20:19
  • $\begingroup$ 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. $\endgroup$ – leo Dec 17 '14 at 20:26
  • $\begingroup$ you are missing k in the numerator. could you please add it...thanks for helping write it nicely $\endgroup$ – leo Dec 17 '14 at 20:29
  • $\begingroup$ I meant you miss k when you wrote the series more managable should not it be (n^k)k/(k+1)! $\endgroup$ – leo Dec 17 '14 at 20:40
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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. $\diamondsuit$

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  • $\begingroup$ Oh, gross. Now a sign error. Two errors down! $\endgroup$ – davidlowryduda Dec 17 '14 at 21:17
  • $\begingroup$ now I am trying to find another example under same power series instead of taking (x_k)=k/k+1 I took (x_k)=1/2^k and I found the result but when I took (x_k)=1/k I could not find the solution and my second question if I take any sequence (x_k) but this is any arbitrarily convergent sequence is there a way to find the sum of power series? $\endgroup$ – leo Dec 18 '14 at 9:45

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