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?


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, heropup, colormegone, Shailesh, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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


Note that we have

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

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.