Need help understanding why my answer is wrong on a direct comparison test.
\[\sum_{0}^{\infty}\frac{ 1 }{ n! }\]
I wonder what you could compare this to...
I saw the n factorial terms as\[\frac{ 1 }{ n }+\frac{ 1 }{ n-1 }+\frac{ 1 }{ n-2 }+.....\frac{ 1 }{ 2 }+\frac{ 1 }{ 1 }\] then compared that to \[\frac{ 1 }{ n }\]which is a divergent p-series. Is it wrong to say that \[\frac{ 1 }{ n } \le \frac{ 1 }{ n! }\]????
Whoa there, where did you get \[\frac{ 1 }{ n }+\frac{ 1 }{ n-1 }+\frac{ 1 }{ n-2 }+.....\frac{ 1 }{ 2 }+\frac{ 1 }{ 1 }\]?
I goofed I am thinking multiplication
So... you found your error? :D
so maybe i Have the concept confused. When I am using the direct comparison test am I comparing the nth partial sums? Or am I comparing the sequence?
The sequence :) Each individual term of the sequence, you compare
so then the 1/n! would be\[\frac{ 1 }{ 0! }+\frac{ 1 }{ 1! }+\frac{ 1 }{ 2! }\] etc?
Yes :D
and I need to find a sequence that diverges that is less than this sequence or one that converges that is greater than this sequence?
Yup.
Yikes!!
It doesn't have to be greater all the time. Just that at some point, it will always be greater/smaller
right but I thought factorials grew faster than exponentials?
In this case, we could use this (convergent) geometric series \[\large \frac1{2^n}\] Now, for n = 0,1,2, and 3 \[\Large \frac1{n!}>\frac1{2^n}\] But for n = 4 onwards \[\large \frac1{n!}<\frac1{2^n}\] So you can use the comparison test there ^.^
I am a little confused, I thought somewhere along the line someone said that factorials grew faster than exponentials.
of course, that's why when factorials are in the DENOMINATOR, they become smaller than when you have EXPONENTIALS at your denominator :)
Duh LOL I was just testing you see if you remembered lol
Is that so? -.-
Thanks that helps. I was just having a slow moment
Join our real-time social learning platform and learn together with your friends!