sequences - monotonic or not


by rohan03
Tags: monotonic, sequences
rohan03
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#1
Apr2-12, 09:21 AM
P: 56
Now I know how this works- but I came across this example and even though I know the answer- the simplification given in the explaination doesn't make sense to me.

the squence is an= {5n/n!}
now applying an+1 and dividing an+1/an
the book indicates = 5/n+1

this is what I don't get how
(5n+1 /(n+1)!)/(5 n/n!) can simplify to that ?

can someone explain please- what am I missing here.
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Number Nine
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#2
Apr2-12, 10:04 AM
P: 771
We have...
[tex]\frac{5^{n+1}}{(n+1)!}\frac{n!}{5^n} = \frac{5\cdot5^{n}}{(n+1)n!}\frac{n!}{5^n}[/tex]
...which very easily simplifies to the expression you provided by cancelling out like terms.
rohan03
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#3
Apr2-12, 11:25 AM
P: 56
right - this is what is not clear to me- I am very new to pure maths
how (n+1)! can be written as - (n+1)n!- may be I am having a dumb moment

micromass
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#4
Apr2-12, 11:29 AM
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sequences - monotonic or not


What is the definition of n! for you?
rohan03
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#5
Apr2-12, 11:32 AM
P: 56
well n! means = any number say 5 then multiplied by 5x4x3x2x1 ( natural numbers in hughest to lowest order)
rohan03
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#6
Apr2-12, 11:34 AM
P: 56
so basically product of positive integres less than or equal to n
micromass
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#7
Apr2-12, 11:35 AM
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So, you have

[tex](n+1)!=(n+1)*n*(n-1)*(n-2)*...*3*2*1[/tex]

Right?

But then we have

[tex](n+1)!=(n+1)*[n*(n-1)*(n-2)*...*3*2*1][/tex]

And the thing in brackets look familiar, no?? Indeed, the bracketed thing is n!
So

[tex](n+1)!=(n+1)*n![/tex]
rohan03
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#8
Apr2-12, 11:39 AM
P: 56
thank you this makes sense- sometimes I just get frustrated with not enough explaination at beiggners level


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