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

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

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

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^ where \(S\) is what we're looking to find.
Looks like you can derive by differentiating \(e^{-x}\).
Ah yeah that's much more efficient.
Oh, \(-13\) should be \(+7\). That fixes everything.
@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.
after cancelling, the index also shifts, so that shouldn't be a problem i think
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.
that issue can be avoided, if we simply shift the index right \[\begin{align*} \frac{1}{e}&=\sum_{n=1}^\infty \frac{(-1)^{n-1}(n^3+1-n^3)}{(n-1)!}\\\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}n^3}{(n-1)!}-\sum_{n=1}^\infty \frac{(-1)^{n-1}(n^3-1)}{(n-1)!}\\\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}n^3}{(n-1)!}-\sum_{n=\color{red}{2}}^\infty \frac{(-1)^{n-1}(n^2+n+1)}{(n-2)!} \end{align*}\]
we can simply shift the index because n=1 produces 0 anyways
\[ e^{-x}=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^n}{n!} \]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 \(n=m+1\):\[ e^{-x}=\sum_{m=0}^{\infty}(-1)^{m}\frac{(m+1)x^m}{m!}= \sum_{n=0}^{\infty}(-1)^{n}\frac{x^n}{n!} \]
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 \(C=2,B=-3\)
now consider $$1+-1+2+-3=-1$$ so we get \(-1/e\) Q.E.D.
wow! how did that work
looks pretty close to @SithsAndGiggles method but also looks more clever!
Definitely less taxing than copy/pasting sums over and over :)
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 :)
$$\begin{align*}e^{-x}&=\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}\\-e^{-x}&=\sum_{n=0}^\infty\frac{(-1)^nnx^{n-1}}{n!}\\e^{-x}&=\sum_{n=0}^\infty\frac{(-1)^nn(n-1)x^{n-2}}{n!}\\-e^{-x}&=\sum_{n=0}^\infty\frac{(-1)^nn(n-1)(n-2)x^{n-3}}{n!}\end{align*}$$so plug in \(x=1\):$$e^{-x}=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\\-e^{-x}=\sum_{n=0}^\infty\frac{(-1)^n n}{n!}\\e^{-x}=\sum_{n=0}^\infty\frac{(-1)^nn(n-1)}{n!}\\-e^{-x}=\sum_{n=0}^\infty\frac{(-1)^n n(n-1)(n-2)}{n!}$$
so it follows if \((n+1)^3=1+3n+2n(n-1)+n(n-1)(n-2)\) then we have: $$\sum_{n=0}^\infty\frac{(-1)^n (n+1)^3}{n!}\\\quad =1\cdot\sum_{n=0}^\infty\frac{(-1)^n}{n!}+3\cdot\sum_{n=0}^\infty\frac{(-1)^nn}{n!}+2\cdot\sum_{n=0}^\infty\frac{(-1)^nn(n-1)}{n!}\\\quad\quad \quad +1\cdot\sum_{n=0}^\infty\frac{(-1)^nn(n-1)(n-2)}{n!}\\\quad=1\cdot e^{-1}+3\cdot(-e^{-1})+2\cdot e^{-1}+1\cdot(-e^{-1})\\\quad =(1-3+2-1)\cdot\frac1e\\\quad=-\frac1e$$
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.
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
@oldrin.bataku this may not affect the solution, but when i solved the system of equations i get \[(n+1)^3= 1+7n+6n(n-1)+n(n-1)(n-2)\]
this wont affect because (1-7+6-1) end up being -1 :)
thats pretty cool actually!! thnks for introducing the special trick xD
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
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
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 !
yep, you just have to find a way to write the polynomial in terms of \((n)_i\) where \((n)_k=n(n-1)(n-2)\cdots(n-k+1)\)
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 $$\begin{align*}1/0!&=1\\7/1!&=7\\12/2!&=6\\6/3!&=1\end{align*}$$
interesting way to pull finite differences into this xD im still trying to make sense why it is giving the coefficients..
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
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
it works because repeated differences: https://en.wikipedia.org/wiki/Finite_difference#Newton.27s_series
if it were not alternating, then i bet the sum is 203e because 1 + 31 + 90 + 65 + 15 + 1 = 203
trivially im right too xD http://www.wolframalpha.com/input/?i=sum_%7Bn%3D0%7D%5Einfty++%28n%2B1%29%5E5%2Fn%21
yep, that is correct
guess il need to review repeated differences, i remember studying finite differences sometime back but don't seem to understand much...
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 $$\frac1e(\tilde B_0+3\tilde B_1+3\tilde B_2+\tilde B_3)=\frac1e(1-2+0+1)=-\frac1e$$
http://mathworld.wolfram.com/ComplementaryBellNumber.html
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
in fact there's an even easier way to do it in terms of complementary Bell numbers: $$\sum_{n=0}^\infty\frac{(-1)^n (n+1)^k}{n!}=-\sum_{n=0}^\infty\frac{(-1)^{n+1}(n+1)^{k+1}}{(n+1)!}=-\tilde B_{k+1}$$
oops, \(-\frac{\tilde B_{k+1}}e\) i mean :-)
@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

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