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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, or this\[\sum_{n=1}^\infty (1)^{n1}\frac{n^3}{(n1)!}\]same difference (or sum, :P)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1wolf gives a pretty answer! https://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%5Cinfty+%281%29%5E%7Bn1%7D%5Cfrac%7Bn%5E3%7D%7B%28n1%29%21%7D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Definitely reminiscent of the power series for \(e^{x}\): \[\sum_{n=0}^\infty (1)^{n}\frac{x^n}{n!}\] Shifting the index, \[e^{x}=\sum_{n=1}^\infty (1)^{n1}\frac{x^{n1}}{(n1)!}\] Let's say \(x=1\), then \[\frac{1}{e}=\sum_{n=1}^\infty \frac{(1)^{n1}}{(n1)!}\] Add and subtract \(n^3\) in the numerator: \[\begin{align*} \frac{1}{e}&=\sum_{n=1}^\infty \frac{(1)^{n1}(n^3+1n^3)}{(n1)!}\\\\ &=\sum_{n=1}^\infty \frac{(1)^{n1}n^3}{(n1)!}\sum_{n=1}^\infty \frac{(1)^{n1}(n^31)}{(n1)!}\\\\ &=\sum_{n=1}^\infty \frac{(1)^{n1}n^3}{(n1)!}\sum_{n=1}^\infty \frac{(1)^{n1}(n^2+n+1)}{(n2)!} \end{align*}\] I'm thinking write \(n^2+n+1\) as a quadratic in \(n2\), so \[n^2+n+1=(n2)^2+5(n2)13\] \[\begin{align*} \frac{1}{e}&=S\sum_{n=1}^\infty \frac{(1)^{n1}((n2)^2+5(n2)13)}{(n2)!}\\\\ &=S\sum_{n=1}^\infty \frac{(1)^{n1}(n2)^2}{(n2)!}5\sum_{n=1}^\infty \frac{(1)^{n1}(n2)}{(n2)!}\\&\quad\quad+13\sum_{n=1}^\infty \frac{(1)^{n1}}{(n2)!}\\\\ &=S\sum_{n=1}^\infty \frac{(1)^{n1}(n2)}{(n3)!}\color{red}{5\sum_{n=1}^\infty \frac{(1)^{n3}}{(n3)!}}\color{blue}{13\sum_{n=1}^\infty \frac{(1)^{n2}}{(n2)!}}\\\\ \frac{1}{e}+\color{red}{\frac{5}{e}}+\color{blue}{\frac{13}{e}}&=S\sum_{n=1}^\infty \frac{(1)^{n1}(n2)}{(n3)!} \end{align*}\] Hmm, there's a mistake somewhere... WA is saying the sum is negative.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0^ where \(S\) is what we're looking to find.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looks like you can derive by differentiating \(e^{x}\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah yeah that's much more efficient.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, \(13\) should be \(+7\). That fixes everything.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0@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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1after cancelling, the index also shifts, so that shouldn't be a problem i think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that issue can be avoided, if we simply shift the index right \[\begin{align*} \frac{1}{e}&=\sum_{n=1}^\infty \frac{(1)^{n1}(n^3+1n^3)}{(n1)!}\\\\ &=\sum_{n=1}^\infty \frac{(1)^{n1}n^3}{(n1)!}\sum_{n=1}^\infty \frac{(1)^{n1}(n^31)}{(n1)!}\\\\ &=\sum_{n=1}^\infty \frac{(1)^{n1}n^3}{(n1)!}\sum_{n=\color{red}{2}}^\infty \frac{(1)^{n1}(n^2+n+1)}{(n2)!} \end{align*}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we can simply shift the index because n=1 produces 0 anyways

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ e^{x}=\sum_{n=0}^{\infty}(1)^{n}\frac{x^n}{n!} \]Differentiate and:\[ e^{x}=\sum_{n=1}^{\infty}(1)^{n}\frac{nx^{n1}}{(n1)!} \]Divide both sides by \(1\) and you get: \[ e^{x}=\sum_{n=1}^{\infty}(1)^{n1}\frac{nx^{n1}}{(n1)!} \]Let \(m = n1\) and so \(n=m+1\):\[ e^{x}=\sum_{m=0}^{\infty}(1)^{m}\frac{(m+1)x^m}{m!}= \sum_{n=0}^{\infty}(1)^{n}\frac{x^n}{n!} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider $$(n+1)^3=ABn+Cn(n1)Dn(n1)(n2)\\n^3+3n^2+3n+1=ABn+C(n^2n)D(n^3+n^2+2n)\\n^3+3n^2+3n+1=A+(BC+2D)n+(C+D)n^2+Dn^3$$so we have \(A=1,D=1\) and \(C=2,B=3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now consider $$1+1+2+3=1$$ so we get \(1/e\) Q.E.D.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1wow! how did that work

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1looks pretty close to @SithsAndGiggles method but also looks more clever!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Definitely less taxing than copy/pasting sums over and over :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1gotcha! 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(n1)+n(n1)(n2) one shot :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\begin{align*}e^{x}&=\sum_{n=0}^\infty\frac{(1)^nx^n}{n!}\\e^{x}&=\sum_{n=0}^\infty\frac{(1)^nnx^{n1}}{n!}\\e^{x}&=\sum_{n=0}^\infty\frac{(1)^nn(n1)x^{n2}}{n!}\\e^{x}&=\sum_{n=0}^\infty\frac{(1)^nn(n1)(n2)x^{n3}}{n!}\end{align*}$$so plug in \(x=1\):$$e^{x}=\sum_{n=0}^\infty\frac{(1)^n}{n!}\\e^{x}=\sum_{n=0}^\infty\frac{(1)^n n}{n!}\\e^{x}=\sum_{n=0}^\infty\frac{(1)^nn(n1)}{n!}\\e^{x}=\sum_{n=0}^\infty\frac{(1)^n n(n1)(n2)}{n!}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it follows if \((n+1)^3=1+3n+2n(n1)+n(n1)(n2)\) then we have: $$\sum_{n=0}^\infty\frac{(1)^n (n+1)^3}{n!}\\\quad =1\cdot\sum_{n=0}^\infty\frac{(1)^n}{n!}+3\cdot\sum_{n=0}^\infty\frac{(1)^nn}{n!}+2\cdot\sum_{n=0}^\infty\frac{(1)^nn(n1)}{n!}\\\quad\quad \quad +1\cdot\sum_{n=0}^\infty\frac{(1)^nn(n1)(n2)}{n!}\\\quad=1\cdot e^{1}+3\cdot(e^{1})+2\cdot e^{1}+1\cdot(e^{1})\\\quad =(13+21)\cdot\frac1e\\\quad=\frac1e$$

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1@oldrin.bataku this may not affect the solution, but when i solved the system of equations i get \[(n+1)^3= 1+7n+6n(n1)+n(n1)(n2)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1this wont affect because (17+61) end up being 1 :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1thats pretty cool actually!! thnks for introducing the special trick xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, you're probably right: $$(n+1)^3=ABn+Cn(n1)Dn(n1)(n2)$$ so we get \(n=0,1,2,3\): $$1=A\\8=AB\\27=A2B+2C\\64=A3B+6C6D$$... which gives \(A=1,B=7,C=6,D=1\), indeed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 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 selfevident when I thought about how to do this problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1guess i can mimic the same to any power https://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%5Cinfty+%281%29%5E%7Bn1%7D%5Cfrac%7Bn%5E5%7D%7B%28n1%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 !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, you just have to find a way to write the polynomial in terms of \((n)_i\) where \((n)_k=n(n1)(n2)\cdots(nk+1)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in 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 $$\begin{align*}1/0!&=1\\7/1!&=7\\12/2!&=6\\6/3!&=1\end{align*}$$

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1interesting way to pull finite differences into this xD im still trying to make sense why it is giving the coefficients..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for \((n+1)^5\) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i'm right: http://www.wolframalpha.com/input/?i=sum_%7Bn%3D0%7D%5Einfty+%281%29%5En+%28n%2B1%29%5E5%2Fn%21

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it works because repeated differences: https://en.wikipedia.org/wiki/Finite_difference#Newton.27s_series

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1if it were not alternating, then i bet the sum is 203e because 1 + 31 + 90 + 65 + 15 + 1 = 203

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1trivially im right too xD http://www.wolframalpha.com/input/?i=sum_%7Bn%3D0%7D%5Einfty++%28n%2B1%29%5E5%2Fn%21

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, that is correct

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1guess il need to review repeated differences, i remember studying finite differences sometime back but don't seem to understand much...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0interestingly 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 $$\frac1e(\tilde B_0+3\tilde B_1+3\tilde B_2+\tilde B_3)=\frac1e(12+0+1)=\frac1e$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in fact there's an even easier way to do it in terms of complementary Bell numbers: $$\sum_{n=0}^\infty\frac{(1)^n (n+1)^k}{n!}=\sum_{n=0}^\infty\frac{(1)^{n+1}(n+1)^{k+1}}{(n+1)!}=\tilde B_{k+1}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, \(\frac{\tilde B_{k+1}}e\) i mean :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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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