sequences - monotonic or not

rohan03

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= {5ⁿ/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
(5ⁿ⁺¹ /(n+1)!)/(5 ⁿ/n!) can simplify to that ?

can someone explain please- what am I missing here.

Number Nine

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

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

What is the definition of n! for you?

rohan03

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

rohan03

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

micromass

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

thank you this makes sense- sometimes I just get frustrated with not enough explaination at beiggners level