anonymous
  • anonymous
True or False & why? (n+1)! = n!(n+1) This question is relating to series.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
yes
ParthKohli
  • ParthKohli
Woah. You confused on here?
ParthKohli
  • ParthKohli
\((n + 1)! = n! \times (n + 1)\)

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More answers

ParthKohli
  • ParthKohli
That's an axiom, isn't it?
anonymous
  • anonymous
n! is a factorial though, in this case n will be infinite. I don't know, I'm asking if it's an axiom lol
ParthKohli
  • ParthKohli
\( \color{Black}{\Rightarrow (4 + 1)! = 4! \times 5 = 1 \times 2 \times 3 \times 4 \times5 = 5!}\)
ParthKohli
  • ParthKohli
That is an axiom, and that's also a way people prove that \(0! = 1\). :)
anonymous
  • anonymous
I suppose that makes sense, it just seems odd to me
vishweshshrimali5
  • vishweshshrimali5
\[n! = n (n-1)(n-2)...\] SO, \[(n+1)! = (n+1)n(n-1)(n-2)... = (n+1)n!\]
anonymous
  • anonymous
\[\sum_{n=1}^{\infty} n!(2x-1)^n\] for example, is one of the easier ones
ParthKohli
  • ParthKohli
\(n! = n(n - 1)(n - 2) \cdots 1\) Correction* :)
anonymous
  • anonymous
At first I couldn't tell what they were doing to get rid of the n!'s on the ratio test
ParthKohli
  • ParthKohli
I believe that a factorial may be expressed as \(\prod\).
vishweshshrimali5
  • vishweshshrimali5
\[THANKS\] @ParthKohli
vishweshshrimali5
  • vishweshshrimali5
\[Correct\]
ParthKohli
  • ParthKohli
\[\prod_{i = 1}^{n}i = n!\]
anonymous
  • anonymous
sometimes we take 0!=1 as a convention :)
anonymous
  • anonymous
Alright so that's the trick, anything else I should know about factoring n! out?
vishweshshrimali5
  • vishweshshrimali5
Well......... Nothing so important.
ParthKohli
  • ParthKohli
It's actually a simple fact.
anonymous
  • anonymous
Simple, but not intuitive at first glance, at least not to me
ParthKohli
  • ParthKohli
\( \color{Black}{\Rightarrow \Large {16! \over 8!} = {16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \cancel{\times 8!} \over \cancel{8!}}}\)
ParthKohli
  • ParthKohli
Let's just express 3! in terms of 2!. \(3! = 3 \times 2 \times 1 = 3 \times (2 \times 1) = 3 \times 2!\)
anonymous
  • anonymous
a small execise : re-write \[\frac{1.3.5.7.........(2n+1)}{2.4.6.8.............2n}\] using "!"
experimentX
  • experimentX
|dw:1343146818151:dw|
experimentX
  • experimentX
|dw:1343146998874:dw|
ParthKohli
  • ParthKohli
Woah! That's some hardcore Mathematics!
anonymous
  • anonymous
Yeah I'm getting an interval of convergence of -0.5 < x < 0.5 And a radius of convergence of 0.5
anonymous
  • anonymous
Would you agree with that @experimentX ? The limit for this series goes to infinity, it diverges.
experimentX
  • experimentX
wait .. something went wrong!!
anonymous
  • anonymous
I'm using the ratio test... I think that's what you did too *looks back at what your wrote*
ParthKohli
  • ParthKohli
What a change! \(\mathbf{Factorials \Longrightarrow Calculus}\)
experimentX
  • experimentX
well .. factorials are the basics!!
experimentX
  • experimentX
lol ... i did the opposite!!
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| \] \[\lim_{n \rightarrow \infty} \left| \frac{(n+1)!(2x-1)^{n+1}}{n!(2x+1)^n} \right| \]
anonymous
  • anonymous
Using the axiom Parth pointed out I can then cancel...
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} \left| \frac{\cancel{n!}(n+1)(2x-1)^{\cancel{n}+1}}{\cancel{n!}\cancel{(2x+1)^n}} \right| = \infty\]
anonymous
  • anonymous
Divergent by the Ratio Test @ParthKohli & @experimentX if 2x-1 = 0 then x \(\neq\) 0.5 So the interval of convergence is only from -0.5 to 0.5, non-inclusive
anonymous
  • anonymous
But the radius? 0.5? 0? o_O
ParthKohli
  • ParthKohli
Oh no! Not THAT type of limits!
anonymous
  • anonymous
PS: People please give @experimentX a medal...
ParthKohli
  • ParthKohli
Done.
experimentX
  • experimentX
|dw:1343147306986:dw| nvm .. i'm at 99, can't grow any further. i guess ratio test is bad test for this series!!
experimentX
  • experimentX
|dw:1343147733699:dw|
anonymous
  • anonymous
Hold on ladies & gents, let me re-ask this specific problem as a new question, that way proper credit can be due and we're not all off-topic technically :-D
experimentX
  • experimentX
no ... i enjoy weird problems!!

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