just made it up....

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List of weirdest questions that have seen....i am all making it up \(\Large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \rm \frac{(n!)^n}{n^{(n!)}}}\) \(\Large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \rm \frac{(n^2+1)!}{n^n}}\)
eh lolz the first converge
should...

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i find it normal though i dont undestand it completly.
eh you made up those series haha let see how the first can converge
let me first cheat with wolf
lol
\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left|\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right|}\)
it is a very good latex practice too
+ i haven't done them in a while... good start to refresh them
no i can't bear writing all that! too lazy
\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left|\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right|}\) \(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}\) btw dit is ratio test
looks doable to me! wolf kicked me out lol didn't give me an answer
I am doing a suicide mission
2nd diverge actually! just limit
i am getting stuck on the first one
\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}\) this is the first outcome of the ratio test a(n+1)/a(n)
you can kill (n!)^n leaving (n+1)^(n+1)
oh...
indeed... i didn't notice that... to complicated it got
(n+1)^(n+1)!=((n+1)^(n+1))^n!
wow.... \(\bf\color{red}{\href{http:///www.wolframalpha.com/input/?i=limit+n%E2%86%92%E2%88%9E+%28%28%28n%2B1%29%21%29%5E%28n%2B1%29n%5E%28n%21%29%29%2F%28%28n%2B1%29%5E%28%28n%2B1%29%21%29%28n%21%29%5En%29}{RESULT ~is~0}}\)
eh really?
yes converges....
i entered the limit, r=0
as n-> inf
in computer program? or you did the limit
i entered in wolfram and did a hiperlink, because the link is kinda too long
http://www.wolframalpha.com/input/?i=limit+n%E2%86%92%E2%88%9E+%28%28%28n%2B1%29%21%29%5E%28n%2B1%29n%5E%28n%21%29%29%2F%28%28n%2B1%29%5E%28%28n%2B1%29%21%29%28n%21%29%5En%29
oh! i think we can do it by hand lolz needs some good cleaning
hello solomon you have the coolest profile pic
can i use it please?
plagiarism.... i would mind that someone looks same as me. do some basic effects to it and change it so that it differs
anyway....
this limit is 0 http://www.wolframalpha.com/input/?i=limit+n%E2%86%92%E2%88%9E+%28n%2F%28n%2B1%29%5E%28n%2B1%29%29%5E%28n%21%29
we can use some taylor to find out but that's too long of work the limit (n+1)^n+1=1
i got disconnected.
yes, that is not something i would right now, i am a bit tired after finals
I will do some more very simple ones I guess. tnx:)
\(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}}\)
\(\large\color{black}{ \displaystyle e^x=\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}\) \(\large\color{black}{ \displaystyle ee^x=e\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}\) \(\large\color{black}{ \displaystyle ee^1=e\sum_{ n=1 }^{ \infty } \frac{1^n}{n!}}\) \(\large\color{black}{ \displaystyle e^2=\sum_{ n=1 }^{ \infty } \frac{e}{n!}}\)
i have an error
this x^n/n! starts from 0
that's pretty good trick :) you can make that start at 1 instead 3
and then \(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2-\frac{e}{0!}-\frac{e}{1!}-\frac{e}{2!} }\) \(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2-2.5e }\)
yes, as long as I change my both sides the same way it can all work out...
\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{n}{n!} }\) for convr/diver
i mean that is obviously convergent
\(\large\color{black}{ \displaystyle (n+1)/(n+1)!~~\times ~~~n!/n}\) \(\large\color{black}{ \displaystyle1/n=0}\)
:)
\(\large\color{black}{ \displaystyle \sum_{ n=k }^{ \infty } \frac{n!}{n^n}}\)
i have seen that one before i guess
\(\large\color{black}{ \displaystyle n!=1\times 2\times 3 \times ... \times n}\) \(\large\color{black}{ \displaystyle n^n=n\times n\times n \times ... \times n}\)
oh,comp test to 2/n^2
ratio works good with that too
\(\large\color{black}{ \displaystyle \frac{n!}{n^n}=\frac{1}{n}\times \frac{2}{n} \times \frac{3}{n} \times \frac{4}{n} \times ~~...~~\times \frac{n}{n}}\) (n times) each of these terms fraction on the right is smaller than 1 besides from 1/n so n!/n^n is in fact smaller than the product of 1st 2 terms
\(\large\color{black}{ \displaystyle \sum_{n=k}^{\infty}\frac{n!}{n^n}=\sum_{n=k}^{\infty}\frac{2}{n^2}}\)
by the way what is the p series 1/n^2 ? 6/pi^2 ?
1/n^p
constant on top doesnot really matter
oh what is it yes pi^2/6
\(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\pi^2/6}\)
and with p=4, pi^4/90
starting from n=1, and w/o coefficients in front
k, lets see what math math got there:)
well you are just comparing there are not equal
you want to find the limit?
they are not equal, but 2/n^2 is larger than n!/n^n
|dw:1435010612413:dw|
|dw:1435010912898:dw|
when you multiply something by a number smaller than 1 you make it smaller
this is why 2/n^2 is greater than n!/n^n
so call n!/n^n A(n) and 2/n^2 B(n)
by definition of comparison test, if B(n) is larger and converges, then A(n) will also converges

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