Question

just made it up....

Answer 1

List of weirdest questions that have seen....i am all making it up


\(\Large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \rm \frac{(n!)^n}{n^{(n!)}}}\)


\(\Large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \rm \frac{(n^2+1)!}{n^n}}\)

Answer 2

eh lolz
the first converge

Answer 3

should...

Answer 4

i find it normal though i dont undestand it completly.

Answer 5

eh you made up those series haha
let see how the first can converge

Answer 6

let me first cheat with wolf

Answer 7

lol

Answer 8

\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left|\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right|}\)

Answer 9

it is a very good latex practice too

Answer 10

+ i haven't done them in a while... good start to refresh them

Answer 11

no i can't bear writing all that! too lazy

Answer 12

\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left|\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right|}\)

\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}\)

btw dit is ratio test

Answer 13

looks doable to me!
wolf kicked me out lol didn't give me an answer

Answer 14

I am doing a suicide mission

Answer 15

2nd diverge actually! just limit

Answer 16

i am getting stuck on the first one

Answer 17

\(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}\)

this is the first outcome of the ratio test a(n+1)/a(n)

Answer 18

you can kill (n!)^n leaving (n+1)^(n+1)

Answer 19

oh...

Answer 20

indeed... i didn't notice that... to complicated it got

Answer 21

(n+1)^(n+1)!=((n+1)^(n+1))^n!

Answer 22

wow....
\(\bf\color{red}{\href{http:///www.wolframalpha.com/input/?i=limit+n%E2%86%92%E2%88%9E+%28%28%28n%2B1%29%21%29%5E%28n%2B1%29n%5E%28n%21%29%29%2F%28%28n%2B1%29%5E%28%28n%2B1%29%21%29%28n%21%29%5En%29}{RESULT ~is~0}}\)

Answer 23

eh really?

Answer 24

yes converges....

Answer 25

i entered the limit, r=0

Answer 26

as n-> inf

Answer 27

in computer program? or you did the limit

Answer 28

i entered in wolfram and did a hiperlink, because the link is kinda too long

Answer 29

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

Answer 30

oh!
i think we can do it by hand lolz
needs some good cleaning

Answer 31

hello solomon you have the coolest profile pic

Answer 32

can i use it please?

Answer 33

plagiarism.... i would mind that someone looks same as me. do some basic effects to it and change it so that it differs

Answer 34

anyway....

Answer 35

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

Answer 36

we can use some taylor to find out but that's too long of work
the limit (n+1)^n+1=1

Answer 37

i got disconnected.

Answer 38

yes, that is not something i would right now, i am a bit tired after finals

Answer 39

I will do some more very simple ones I guess. tnx:)

Answer 40

\(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}}\)

Answer 41

\(\large\color{black}{ \displaystyle e^x=\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}\)

\(\large\color{black}{ \displaystyle ee^x=e\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}\)

\(\large\color{black}{ \displaystyle ee^1=e\sum_{ n=1 }^{ \infty } \frac{1^n}{n!}}\)

\(\large\color{black}{ \displaystyle e^2=\sum_{ n=1 }^{ \infty } \frac{e}{n!}}\)

Answer 42

i have an error

Answer 43

this x^n/n! starts from 0

Answer 44

that's pretty good trick :)
you can make that start at 1 instead 3

Answer 45

and then \(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2-\frac{e}{0!}-\frac{e}{1!}-\frac{e}{2!} }\)

\(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2-2.5e }\)

Answer 46

yes, as long as I change my both sides the same way it can all work out...

Answer 47

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{n}{n!} }\) for convr/diver

Answer 48

i mean that is obviously convergent

Answer 49

\(\large\color{black}{ \displaystyle (n+1)/(n+1)!~~\times ~~~n!/n}\)
\(\large\color{black}{ \displaystyle1/n=0}\)

Answer 50

:)

Answer 51

\(\large\color{black}{ \displaystyle \sum_{ n=k }^{ \infty } \frac{n!}{n^n}}\)

Answer 52

i have seen that one before i guess

Answer 53

\(\large\color{black}{ \displaystyle n!=1\times 2\times 3 \times ... \times n}\)
\(\large\color{black}{ \displaystyle n^n=n\times n\times n \times ... \times n}\)

Answer 54

oh,comp test to 2/n^2

Answer 55

ratio works good with that too

Answer 56

\(\large\color{black}{ \displaystyle \frac{n!}{n^n}=\frac{1}{n}\times \frac{2}{n} \times \frac{3}{n} \times \frac{4}{n} \times ~~...~~\times \frac{n}{n}}\)

(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

Answer 57

\(\large\color{black}{ \displaystyle \sum_{n=k}^{\infty}\frac{n!}{n^n}=\sum_{n=k}^{\infty}\frac{2}{n^2}}\)

Answer 58

by the way what is the p series 1/n^2 ? 6/pi^2 ?

Answer 59

1/n^p

Answer 60

constant on top doesnot really matter

Answer 61

oh what is it yes pi^2/6

Answer 62

\(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\pi^2/6}\)

Answer 63

and with p=4, pi^4/90

Answer 64

starting from n=1, and w/o coefficients in front

Answer 65

k, lets see what math math got there:)

Answer 66

well you are just comparing there are not equal

Answer 67

you want to find the limit?

Answer 68

they are not equal, but 2/n^2 is larger than n!/n^n