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Sequences  monotonic or not 
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#1
Apr212, 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 a_{n}= {5^{n}/n!} now applying a_{n+1} and dividing a_{n+1}/a_{n} the book indicates = 5/n+1 this is what I don't get how (5^{n+1} /(n+1)!)/(5 ^{n}/n!) can simplify to that ? can someone explain please what am I missing here. 


#2
Apr212, 10:04 AM

P: 772

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. 


#3
Apr212, 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 


#4
Apr212, 11:29 AM

Mentor
P: 18,086

Sequences  monotonic or not
What is the definition of n! for you?



#5
Apr212, 11:32 AM

P: 56

well n! means = any number say 5 then multiplied by 5x4x3x2x1 ( natural numbers in hughest to lowest order)



#6
Apr212, 11:34 AM

P: 56

so basically product of positive integres less than or equal to n



#7
Apr212, 11:35 AM

Mentor
P: 18,086

So, you have
[tex](n+1)!=(n+1)*n*(n1)*(n2)*...*3*2*1[/tex] Right? But then we have [tex](n+1)!=(n+1)*[n*(n1)*(n2)*...*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] 


#8
Apr212, 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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