## ganeshie8 one year ago Evaluate the sum $\large 1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$

1. anonymous

Right, or this$\sum_{n=1}^\infty (-1)^{n-1}\frac{n^3}{(n-1)!}$same difference (or sum, :P)

2. ganeshie8
3. anonymous

Definitely reminiscent of the power series for $$e^{-x}$$: $\sum_{n=0}^\infty (-1)^{n}\frac{x^n}{n!}$ Shifting the index, $e^{-x}=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^{n-1}}{(n-1)!}$ Let's say $$x=1$$, then $\frac{1}{e}=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n-1)!}$ Add and subtract $$n^3$$ in the numerator: \begin{align*} \frac{1}{e}&=\sum_{n=1}^\infty \frac{(-1)^{n-1}(n^3+1-n^3)}{(n-1)!}\\\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}n^3}{(n-1)!}-\sum_{n=1}^\infty \frac{(-1)^{n-1}(n^3-1)}{(n-1)!}\\\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}n^3}{(n-1)!}-\sum_{n=1}^\infty \frac{(-1)^{n-1}(n^2+n+1)}{(n-2)!} \end{align*} I'm thinking write $$n^2+n+1$$ as a quadratic in $$n-2$$, so $n^2+n+1=(n-2)^2+5(n-2)-13$ \begin{align*} \frac{1}{e}&=S-\sum_{n=1}^\infty \frac{(-1)^{n-1}((n-2)^2+5(n-2)-13)}{(n-2)!}\\\\ &=S-\sum_{n=1}^\infty \frac{(-1)^{n-1}(n-2)^2}{(n-2)!}-5\sum_{n=1}^\infty \frac{(-1)^{n-1}(n-2)}{(n-2)!}\\&\quad\quad+13\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n-2)!}\\\\ &=S-\sum_{n=1}^\infty \frac{(-1)^{n-1}(n-2)}{(n-3)!}-\color{red}{5\sum_{n=1}^\infty \frac{(-1)^{n-3}}{(n-3)!}}-\color{blue}{13\sum_{n=1}^\infty \frac{(-1)^{n-2}}{(n-2)!}}\\\\ \frac{1}{e}+\color{red}{\frac{5}{e}}+\color{blue}{\frac{13}{e}}&=S-\sum_{n=1}^\infty \frac{(-1)^{n-1}(n-2)}{(n-3)!} \end{align*} Hmm, there's a mistake somewhere... W|A is saying the sum is negative.

4. anonymous

^ where $$S$$ is what we're looking to find.

5. anonymous

Looks like you can derive by differentiating $$e^{-x}$$.

6. anonymous

Ah yeah that's much more efficient.

7. anonymous

Oh, $$-13$$ should be $$+7$$. That fixes everything.

8. geerky42

@SmthsAndGiggles I think few series are problematic? For example, for series in last step, just check first term n=1, and you would have negative fractional (-2)! in denominator.

9. ganeshie8

after cancelling, the index also shifts, so that shouldn't be a problem i think

10. anonymous

I recall seeing $$n!$$ defined to be $$0$$ if $$n$$ is negative in some contexts, so we could use that to our advantage, but that's kind of a cheap trick.

11. ganeshie8

that issue can be avoided, if we simply shift the index right \begin{align*} \frac{1}{e}&=\sum_{n=1}^\infty \frac{(-1)^{n-1}(n^3+1-n^3)}{(n-1)!}\\\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}n^3}{(n-1)!}-\sum_{n=1}^\infty \frac{(-1)^{n-1}(n^3-1)}{(n-1)!}\\\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}n^3}{(n-1)!}-\sum_{n=\color{red}{2}}^\infty \frac{(-1)^{n-1}(n^2+n+1)}{(n-2)!} \end{align*}

12. ganeshie8

we can simply shift the index because n=1 produces 0 anyways

13. anonymous

$e^{-x}=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^n}{n!}$Differentiate and:$-e^{-x}=\sum_{n=1}^{\infty}(-1)^{n}\frac{nx^{n-1}}{(n-1)!}$Divide both sides by $$-1$$ and you get: $e^{-x}=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{nx^{n-1}}{(n-1)!}$Let $$m = n-1$$ and so $$n=m+1$$:$e^{-x}=\sum_{m=0}^{\infty}(-1)^{m}\frac{(m+1)x^m}{m!}= \sum_{n=0}^{\infty}(-1)^{n}\frac{x^n}{n!}$

14. anonymous

consider $$(n+1)^3=A-Bn+Cn(n-1)-Dn(n-1)(n-2)\\n^3+3n^2+3n+1=A-Bn+C(n^2-n)-D(n^3+n^2+2n)\\n^3+3n^2+3n+1=A+(B-C+2D)n+(C+D)n^2+Dn^3$$so we have $$A=1,D=-1$$ and $$C=2,B=-3$$

15. anonymous

now consider $$1+-1+2+-3=-1$$ so we get $$-1/e$$ Q.E.D.

16. ganeshie8

wow! how did that work

17. ganeshie8

looks pretty close to @SithsAndGiggles method but also looks more clever!

18. anonymous

Definitely less taxing than copy/pasting sums over and over :)

19. ganeshie8

gotcha! both your methods are identical, its only that oldrin.bataku has managed to keep it in short form by writing (n+1)^3 as 1+7n+6n(n-1)+n(n-1)(n-2) one shot :)

20. anonymous

\begin{align*}e^{-x}&=\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}\\-e^{-x}&=\sum_{n=0}^\infty\frac{(-1)^nnx^{n-1}}{n!}\\e^{-x}&=\sum_{n=0}^\infty\frac{(-1)^nn(n-1)x^{n-2}}{n!}\\-e^{-x}&=\sum_{n=0}^\infty\frac{(-1)^nn(n-1)(n-2)x^{n-3}}{n!}\end{align*}so plug in $$x=1$$:$$e^{-x}=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\\-e^{-x}=\sum_{n=0}^\infty\frac{(-1)^n n}{n!}\\e^{-x}=\sum_{n=0}^\infty\frac{(-1)^nn(n-1)}{n!}\\-e^{-x}=\sum_{n=0}^\infty\frac{(-1)^n n(n-1)(n-2)}{n!}$$

21. anonymous

so it follows if $$(n+1)^3=1+3n+2n(n-1)+n(n-1)(n-2)$$ then we have: $$\sum_{n=0}^\infty\frac{(-1)^n (n+1)^3}{n!}\\\quad =1\cdot\sum_{n=0}^\infty\frac{(-1)^n}{n!}+3\cdot\sum_{n=0}^\infty\frac{(-1)^nn}{n!}+2\cdot\sum_{n=0}^\infty\frac{(-1)^nn(n-1)}{n!}\\\quad\quad \quad +1\cdot\sum_{n=0}^\infty\frac{(-1)^nn(n-1)(n-2)}{n!}\\\quad=1\cdot e^{-1}+3\cdot(-e^{-1})+2\cdot e^{-1}+1\cdot(-e^{-1})\\\quad =(1-3+2-1)\cdot\frac1e\\\quad=-\frac1e$$

22. geerky42

You know how an Italian chef kisses his fingers and says something in Italian that translates to "A masterpiece" after tasting their own dish? That's how I feel about this.

23. anonymous

I didn't need to write it out, though, because the behavior of $$e^x$$ under differentiation is easy enough to see in your head and I recognized that derivatives $$e^{-x}$$ alternate in sign; then it just needed $$x=1$$ -- not that bad

24. ganeshie8

@oldrin.bataku this may not affect the solution, but when i solved the system of equations i get $(n+1)^3= 1+7n+6n(n-1)+n(n-1)(n-2)$

25. ganeshie8

this wont affect because (1-7+6-1) end up being -1 :)

26. ganeshie8

thats pretty cool actually!! thnks for introducing the special trick xD

27. anonymous

oops, you're probably right: $$(n+1)^3=A-Bn+Cn(n-1)-Dn(n-1)(n-2)$$ so we get $$n=0,1,2,3$$: $$1=A\\8=A-B\\27=A-2B+2C\\64=A-3B+6C-6D$$... which gives $$A=1,B=-7,C=6,D=-1$$, indeed

28. anonymous

but yeah, I'm not sure if it's a popular trick of any sort as I've never seen it before, it just seemed kinda self-evident when I thought about how to do this problem

29. ganeshie8

guess i can mimic the same to any power https://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%5Cinfty+%28-1%29%5E%7Bn-1%7D%5Cfrac%7Bn%5E5%7D%7B%28n-1%29%21%7D just need to write (n+1)^5 as linear combinations of earlier products and add/subtract 1/e's ! im loving this method !

30. anonymous

yep, you just have to find a way to write the polynomial in terms of $$(n)_i$$ where $$(n)_k=n(n-1)(n-2)\cdots(n-k+1)$$

31. anonymous

in fact, I bet you can use divided differences to find the coefficients: consider differences of (n+1)^3 for $$n=0,1,2,3$$: 1 7 8 12 19 6 27 18 37 64 so our coefficients are \begin{align*}1/0!&=1\\7/1!&=7\\12/2!&=6\\6/3!&=1\end{align*}

32. ganeshie8

interesting way to pull finite differences into this xD im still trying to make sense why it is giving the coefficients..

33. anonymous

for $$(n+1)^5$$ we have for $$n=0,1,2,3,4,5$$ 1 31 32 180 211 390 243 570 360 781 750 120 1024 1320 480 2101 1230 3125 2550 4651 7776 so 1, 31, 90, 65, 15, 1 so 1 - 31 + 90 - 65 + 15 - 1 = 9 so i predict the sum should be 9/e

34. anonymous
35. anonymous

it works because repeated differences: https://en.wikipedia.org/wiki/Finite_difference#Newton.27s_series

36. ganeshie8

if it were not alternating, then i bet the sum is 203e because 1 + 31 + 90 + 65 + 15 + 1 = 203

37. ganeshie8

trivially im right too xD http://www.wolframalpha.com/input/?i=sum_%7Bn%3D0%7D%5Einfty++%28n%2B1%29%5E5%2Fn%21

38. anonymous

yep, that is correct

39. ganeshie8

guess il need to review repeated differences, i remember studying finite differences sometime back but don't seem to understand much...

40. anonymous

interestingly enough, there's also way to do it when expanding in terms of complementary Bell numbers: $$\frac1e\tilde B_n=\sum_{k=0}^\infty\frac{(-1)^n k^n}{k!}$$ so we observe for $$(n+1)^3=1+3n+3n^2+n^3$$ gives us $$\frac1e(\tilde B_0+3\tilde B_1+3\tilde B_2+\tilde B_3)=\frac1e(1-2+0+1)=-\frac1e$$

41. anonymous
42. anonymous

the coefficients for our expansion in terms of falling factorials is actually just the Stirling numbers of the second kind: http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html see 1,7,6,1

43. anonymous

in fact there's an even easier way to do it in terms of complementary Bell numbers: $$\sum_{n=0}^\infty\frac{(-1)^n (n+1)^k}{n!}=-\sum_{n=0}^\infty\frac{(-1)^{n+1}(n+1)^{k+1}}{(n+1)!}=-\tilde B_{k+1}$$

44. anonymous

oops, $$-\frac{\tilde B_{k+1}}e$$ i mean :-)

45. anonymous

@ganeshie8 here, this is why repeated differences works -- falling factorials obey a rule like the power rule of differentiation: https://en.wikipedia.org/wiki/Pochhammer_symbol#Relation_to_umbral_calculus