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I'm trying to prove this equality.

$$\sum_{k=0}^\infty\frac{1}{k!}\frac{x^{k+6}}{(k+6)} {=} e^x(x^5-5x^4+20x^3-60x^2+120x-120)+120$$

posted by: https://math.stackexchange.com/q/832368.

How do I get the final result from the sum above? I mean, to prove it I should differentiate and then integrate, but then I should start again with the main problem. So, is there another method to get that result?

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closed as off-topic by Did, heropup, colormegone, Shailesh, JonMark Perry Jun 1 '16 at 1:31

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  • $\begingroup$ No, you can't even do what you wrote. You cannot pull out $\sum_{k=0}^\infty \frac{x^k}{k!}$ from the summand. To see why it obviously fails, try comparing the LHS for $x = 2$, which will converge; yet the RHS of what you wrote diverges for $|x| > 1$. $\endgroup$ – heropup May 31 '16 at 17:57
  • $\begingroup$ Any further hint to go on? $\endgroup$ – Lorenzo Comoglio May 31 '16 at 17:58
  • $\begingroup$ try separating the sum and re-indexing $\endgroup$ – danimal May 31 '16 at 18:01
  • $\begingroup$ math.stackexchange.com/q/832368 $\endgroup$ – Lorenzo Comoglio May 31 '16 at 18:08
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Define $$f_m(x) = \sum_{k=0}^\infty \frac{x^{k+m}}{k!(k+m)}, \quad m \in \mathbb Z^+.$$ Then $$f'_m(x) = \sum_{k=0}^\infty \frac{x^{k+m-1}}{k!} = x^{m-1} \sum_{k=0}^\infty \frac{x^k}{k!} = x^{m-1} e^x.$$ Since $f_m(0) = 0$, we conclude by the Fundamental Theorem of calculus $$f_m(x) = \int_{t=0}^x f'_m(t) \, dt = \int_{t=0}^x t^{m-1} e^t \, dt.$$ Now choose $m = 6$ and perfom the requisite integration by parts.

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  • $\begingroup$ I'm trying to prove this: math.stackexchange.com/q/832368 Why should I integrate another time? $\endgroup$ – Lorenzo Comoglio May 31 '16 at 18:05
  • $\begingroup$ @LorenzoComoglio This is why it is important to provide proper context whenever you ask a question. Why did you wait until after you received answers to mention that you were looking at another question? $\endgroup$ – heropup May 31 '16 at 18:18
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HINT:

Note that we have

$$\begin{align} \sum_{k=0}^\infty \frac{x^{k+6}}{k!(k+6)}&=\sum_{k=0}^\infty \frac{1}{k!}\int_0^x t^{k+5}\,dt\\\\ &=\int_0^x t^5 \sum_{k=0}^\infty \frac{t^k}{k!}\,dx\\\\ &=\int_0^x t^5 e^t\,dt \end{align}$$

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  • $\begingroup$ Thing is, math.stackexchange.com/q/832368 we're coming back to the main problem. $\endgroup$ – Lorenzo Comoglio May 31 '16 at 18:12
  • $\begingroup$ Yes, have you used Feynmann's Trick? See Felix Martin's solution. It is a very efficient approach. $\endgroup$ – Mark Viola May 31 '16 at 18:12
  • $\begingroup$ Dr MV, how do I solve: $\int x^5 e^xdx$ this way, but without coming back to the first problem... I mean, we have the equality stated in the question, and to solve it we have to do integration by parts? The question is to solve $\sum_{k=0}^\infty\frac{1}{k!}\frac{x^{k+6}}{(k+6)} {=} e^x(x^5-5x^4+20x^3-60x^2+120x-120)+120$ without coming back to integration by parts... $\endgroup$ – Lorenzo Comoglio May 31 '16 at 18:17

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