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

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ParthKohliBest ResponseYou've already chosen the best response.2
Woah. You confused on here?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
\((n + 1)! = n! \times (n + 1)\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
That's an axiom, isn't it?
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
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
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
\( \color{Black}{\Rightarrow (4 + 1)! = 4! \times 5 = 1 \times 2 \times 3 \times 4 \times5 = 5!}\)
 one year ago

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

agentx5Best ResponseYou've already chosen the best response.0
I suppose that makes sense, it just seems odd to me
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.0
\[n! = n (n1)(n2)...\] SO, \[(n+1)! = (n+1)n(n1)(n2)... = (n+1)n!\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
\[\sum_{n=1}^{\infty} n!(2x1)^n\] for example, is one of the easier ones
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
\(n! = n(n  1)(n  2) \cdots 1\) Correction* :)
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
At first I couldn't tell what they were doing to get rid of the n!'s on the ratio test
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
I believe that a factorial may be expressed as \(\prod\).
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.0
\[THANKS\] @ParthKohli
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
\[\prod_{i = 1}^{n}i = n!\]
 one year ago

NeemoBest ResponseYou've already chosen the best response.0
sometimes we take 0!=1 as a convention :)
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Alright so that's the trick, anything else I should know about factoring n! out?
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.0
Well......... Nothing so important.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
It's actually a simple fact.
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Simple, but not intuitive at first glance, at least not to me
 one year ago

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

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

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

experimentXBest ResponseYou've already chosen the best response.1
dw:1343146818151:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1343146998874:dw
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
Woah! That's some hardcore Mathematics!
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Yeah I'm getting an interval of convergence of 0.5 < x < 0.5 And a radius of convergence of 0.5
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Would you agree with that @experimentX ? The limit for this series goes to infinity, it diverges.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
wait .. something went wrong!!
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
I'm using the ratio test... I think that's what you did too *looks back at what your wrote*
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
What a change! \(\mathbf{Factorials \Longrightarrow Calculus}\)
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
well .. factorials are the basics!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
lol ... i did the opposite!!
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
\[\lim_{n \rightarrow \infty} \left \frac{a_{n+1}}{a_n} \right \] \[\lim_{n \rightarrow \infty} \left \frac{(n+1)!(2x1)^{n+1}}{n!(2x+1)^n} \right \]
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Using the axiom Parth pointed out I can then cancel...
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
\[\lim_{n \rightarrow \infty} \left \frac{\cancel{n!}(n+1)(2x1)^{\cancel{n}+1}}{\cancel{n!}\cancel{(2x+1)^n}} \right = \infty\]
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Divergent 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
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
But the radius? 0.5? 0? o_O
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
Oh no! Not THAT type of limits!
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
PS: People please give @experimentX a medal...
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
dw:1343147733699:dw
 one year ago

agentx5Best ResponseYou've already chosen the best response.0
Hold 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
no ... i enjoy weird problems!!
 one year ago
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