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anonymous
 3 years ago
True or False & why?
(n+1)! = n!(n+1)
This question is relating to series.
anonymous
 3 years ago
True or False & why? (n+1)! = n!(n+1) This question is relating to series.

This Question is Closed

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2Woah. You confused on here?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2\((n + 1)! = n! \times (n + 1)\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2That's an axiom, isn't it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0n! 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
 3 years ago
Best ResponseYou've already chosen the best response.2\( \color{Black}{\Rightarrow (4 + 1)! = 4! \times 5 = 1 \times 2 \times 3 \times 4 \times5 = 5!}\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2That is an axiom, and that's also a way people prove that \(0! = 1\). :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I suppose that makes sense, it just seems odd to me

vishweshshrimali5
 3 years ago
Best ResponseYou've already chosen the best response.0\[n! = n (n1)(n2)...\] SO, \[(n+1)! = (n+1)n(n1)(n2)... = (n+1)n!\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} n!(2x1)^n\] for example, is one of the easier ones

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2\(n! = n(n  1)(n  2) \cdots 1\) Correction* :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0At first I couldn't tell what they were doing to get rid of the n!'s on the ratio test

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2I believe that a factorial may be expressed as \(\prod\).

vishweshshrimali5
 3 years ago
Best ResponseYou've already chosen the best response.0\[THANKS\] @ParthKohli

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2\[\prod_{i = 1}^{n}i = n!\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sometimes we take 0!=1 as a convention :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Alright so that's the trick, anything else I should know about factoring n! out?

vishweshshrimali5
 3 years ago
Best ResponseYou've already chosen the best response.0Well......... Nothing so important.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2It's actually a simple fact.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Simple, but not intuitive at first glance, at least not to me

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2\( \color{Black}{\Rightarrow \Large {16! \over 8!} = {16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \cancel{\times 8!} \over \cancel{8!}}}\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2Let's just express 3! in terms of 2!. \(3! = 3 \times 2 \times 1 = 3 \times (2 \times 1) = 3 \times 2!\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a small execise : rewrite \[\frac{1.3.5.7.........(2n+1)}{2.4.6.8.............2n}\] using "!"

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343146818151:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343146998874:dw

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2Woah! That's some hardcore Mathematics!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah I'm getting an interval of convergence of 0.5 < x < 0.5 And a radius of convergence of 0.5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Would you agree with that @experimentX ? The limit for this series goes to infinity, it diverges.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1wait .. something went wrong!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm using the ratio test... I think that's what you did too *looks back at what your wrote*

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2What a change! \(\mathbf{Factorials \Longrightarrow Calculus}\)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1well .. factorials are the basics!!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1lol ... i did the opposite!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \left \frac{a_{n+1}}{a_n} \right \] \[\lim_{n \rightarrow \infty} \left \frac{(n+1)!(2x1)^{n+1}}{n!(2x+1)^n} \right \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Using the axiom Parth pointed out I can then cancel...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \left \frac{\cancel{n!}(n+1)(2x1)^{\cancel{n}+1}}{\cancel{n!}\cancel{(2x+1)^n}} \right = \infty\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Divergent by the Ratio Test @ParthKohli & @experimentX if 2x1 = 0 then x \(\neq\) 0.5 So the interval of convergence is only from 0.5 to 0.5, noninclusive

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But the radius? 0.5? 0? o_O

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.2Oh no! Not THAT type of limits!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0PS: People please give @experimentX a medal...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343147306986:dw nvm .. i'm at 99, can't grow any further. i guess ratio test is bad test for this series!!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343147733699:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hold on ladies & gents, let me reask this specific problem as a new question, that way proper credit can be due and we're not all offtopic technically :D

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1no ... i enjoy weird problems!!
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