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SolomonZelman
 one year ago
just made it up....
SolomonZelman
 one year ago
just made it up....

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1List of weirdest questions that have seen....i am all making it up \(\Large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \rm \frac{(n!)^n}{n^{(n!)}}}\) \(\Large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \rm \frac{(n^2+1)!}{n^n}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1eh lolz the first converge

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i find it normal though i dont undestand it completly.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1eh you made up those series haha let see how the first can converge

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1let me first cheat with wolf

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1it is a very good latex practice too

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1+ i haven't done them in a while... good start to refresh them

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1no i can't bear writing all that! too lazy

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right}\) \(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}\) btw dit is ratio test

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1looks doable to me! wolf kicked me out lol didn't give me an answer

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I am doing a suicide mission

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.12nd diverge actually! just limit

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1i am getting stuck on the first one

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}\) this is the first outcome of the ratio test a(n+1)/a(n)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1you can kill (n!)^n leaving (n+1)^(n+1)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1indeed... i didn't notice that... to complicated it got

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1(n+1)^(n+1)!=((n+1)^(n+1))^n!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1wow.... \(\bf\color{red}{\href{http:///www.wolframalpha.com/input/?i=limit+n%E2%86%92%E2%88%9E+%28%28%28n%2B1%29%21%29%5E%28n%2B1%29n%5E%28n%21%29%29%2F%28%28n%2B1%29%5E%28%28n%2B1%29%21%29%28n%21%29%5En%29}{RESULT ~is~0}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1eh really?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes converges....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1i entered the limit, r=0

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1in computer program? or you did the limit

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1i entered in wolfram and did a hiperlink, because the link is kinda too long

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh! i think we can do it by hand lolz needs some good cleaning

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hello solomon you have the coolest profile pic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can i use it please?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1plagiarism.... i would mind that someone looks same as me. do some basic effects to it and change it so that it differs

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1this 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

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1we can use some taylor to find out but that's too long of work the limit (n+1)^n+1=1

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1i got disconnected.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, that is not something i would right now, i am a bit tired after finals

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I will do some more very simple ones I guess. tnx:)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle e^x=\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}\) \(\large\color{black}{ \displaystyle ee^x=e\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}\) \(\large\color{black}{ \displaystyle ee^1=e\sum_{ n=1 }^{ \infty } \frac{1^n}{n!}}\) \(\large\color{black}{ \displaystyle e^2=\sum_{ n=1 }^{ \infty } \frac{e}{n!}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1this x^n/n! starts from 0

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1that's pretty good trick :) you can make that start at 1 instead 3

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and then \(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2\frac{e}{0!}\frac{e}{1!}\frac{e}{2!} }\) \(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^22.5e }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, as long as I change my both sides the same way it can all work out...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{n}{n!} }\) for convr/diver

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1i mean that is obviously convergent

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle (n+1)/(n+1)!~~\times ~~~n!/n}\) \(\large\color{black}{ \displaystyle1/n=0}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \sum_{ n=k }^{ \infty } \frac{n!}{n^n}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1i have seen that one before i guess

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle n!=1\times 2\times 3 \times ... \times n}\) \(\large\color{black}{ \displaystyle n^n=n\times n\times n \times ... \times n}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh,comp test to 2/n^2

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1ratio works good with that too

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \frac{n!}{n^n}=\frac{1}{n}\times \frac{2}{n} \times \frac{3}{n} \times \frac{4}{n} \times ~~...~~\times \frac{n}{n}}\) (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

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \sum_{n=k}^{\infty}\frac{n!}{n^n}=\sum_{n=k}^{\infty}\frac{2}{n^2}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1by the way what is the p series 1/n^2 ? 6/pi^2 ?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1constant on top doesnot really matter

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh what is it yes pi^2/6

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\pi^2/6}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and with p=4, pi^4/90

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1starting from n=1, and w/o coefficients in front

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1k, lets see what math math got there:)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1well you are just comparing there are not equal

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1you want to find the limit?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1they are not equal, but 2/n^2 is larger than n!/n^n

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435010612413:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435010912898:dw

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1when you multiply something by a number smaller than 1 you make it smaller

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1this is why 2/n^2 is greater than n!/n^n

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1so call n!/n^n A(n) and 2/n^2 B(n)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1by definition of comparison test, if B(n) is larger and converges, then A(n) will also converges
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