Question

Evaluate the sum
\[\large 1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots\]

Answer 1

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

Answer 2

wolf gives a pretty answer!
https://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%5Cinfty+%28-1%29%5E%7Bn-1%7D%5Cfrac%7Bn%5E3%7D%7B%28n-1%29%21%7D

Answer 3

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.

Answer 4

^ where \(S\) is what we're looking to find.

Answer 5

Looks like you can derive by differentiating \(e^{-x}\).

Answer 6

Ah yeah that's much more efficient.

Answer 7

Oh, \(-13\) should be \(+7\). That fixes everything.

Answer 8

@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.

Answer 9

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

Answer 10

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.

Answer 11

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*}\]

Answer 12

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

Answer 13

\[
e^{-x}=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^n}{n!}
\]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 \(n=m+1\):\[
e^{-x}=\sum_{m=0}^{\infty}(-1)^{m}\frac{(m+1)x^m}{m!}= \sum_{n=0}^{\infty}(-1)^{n}\frac{x^n}{n!}
\]

Answer 14

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 \(C=2,B=-3\)

Answer 15

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

Answer 16

wow! how did that work

Answer 17

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

Answer 18

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

Answer 19

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 :)

Answer 20

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

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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 !

Answer 30

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

Answer 31

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 $$\begin{align*}1/0!&=1\\7/1!&=7\\12/2!&=6\\6/3!&=1\end{align*}$$

Answer 32

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

Answer 33

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

Answer 34

and i'm right: http://www.wolframalpha.com/input/?i=sum_%7Bn%3D0%7D%5Einfty+%28-1%29%5En+%28n%2B1%29%5E5%2Fn%21

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

yep, that is correct

Answer 39

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

Answer 40

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 $$\frac1e(\tilde B_0+3\tilde B_1+3\tilde B_2+\tilde B_3)=\frac1e(1-2+0+1)=-\frac1e$$

Answer 41

http://mathworld.wolfram.com/ComplementaryBellNumber.html

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

@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