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

1. SolomonZelman

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

2. xapproachesinfinity

eh lolz the first converge

3. SolomonZelman

should...

4. mathmath333

i find it normal though i dont undestand it completly.

5. xapproachesinfinity

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

6. xapproachesinfinity

let me first cheat with wolf

7. mathmath333

lol

8. SolomonZelman

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

9. SolomonZelman

it is a very good latex practice too

10. SolomonZelman

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

11. xapproachesinfinity

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

12. SolomonZelman

$$\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left|\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right|}$$ $$\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}$$ btw dit is ratio test

13. xapproachesinfinity

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

14. SolomonZelman

I am doing a suicide mission

15. xapproachesinfinity

2nd diverge actually! just limit

16. SolomonZelman

i am getting stuck on the first one

17. SolomonZelman

$$\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}$$ this is the first outcome of the ratio test a(n+1)/a(n)

18. xapproachesinfinity

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

19. SolomonZelman

oh...

20. SolomonZelman

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

21. xapproachesinfinity

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

22. SolomonZelman

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

23. xapproachesinfinity

eh really?

24. SolomonZelman

yes converges....

25. SolomonZelman

i entered the limit, r=0

26. SolomonZelman

as n-> inf

27. xapproachesinfinity

in computer program? or you did the limit

28. SolomonZelman

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

29. SolomonZelman
30. xapproachesinfinity

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

31. anonymous

hello solomon you have the coolest profile pic

32. anonymous

33. SolomonZelman

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

34. SolomonZelman

anyway....

35. xapproachesinfinity
36. xapproachesinfinity

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

37. SolomonZelman

i got disconnected.

38. SolomonZelman

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

39. SolomonZelman

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

40. SolomonZelman

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

41. SolomonZelman

$$\large\color{black}{ \displaystyle e^x=\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}$$ $$\large\color{black}{ \displaystyle ee^x=e\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}$$ $$\large\color{black}{ \displaystyle ee^1=e\sum_{ n=1 }^{ \infty } \frac{1^n}{n!}}$$ $$\large\color{black}{ \displaystyle e^2=\sum_{ n=1 }^{ \infty } \frac{e}{n!}}$$

42. SolomonZelman

i have an error

43. SolomonZelman

this x^n/n! starts from 0

44. xapproachesinfinity

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

45. SolomonZelman

and then $$\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2-\frac{e}{0!}-\frac{e}{1!}-\frac{e}{2!} }$$ $$\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2-2.5e }$$

46. SolomonZelman

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

47. SolomonZelman

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

48. SolomonZelman

i mean that is obviously convergent

49. SolomonZelman

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

50. SolomonZelman

:)

51. SolomonZelman

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

52. xapproachesinfinity

i have seen that one before i guess

53. SolomonZelman

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

54. SolomonZelman

oh,comp test to 2/n^2

55. xapproachesinfinity

ratio works good with that too

56. SolomonZelman

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

57. SolomonZelman

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

58. SolomonZelman

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

59. xapproachesinfinity

1/n^p

60. xapproachesinfinity

constant on top doesnot really matter

61. xapproachesinfinity

oh what is it yes pi^2/6

62. SolomonZelman

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

63. SolomonZelman

and with p=4, pi^4/90

64. SolomonZelman

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

65. SolomonZelman

k, lets see what math math got there:)

66. xapproachesinfinity

well you are just comparing there are not equal

67. xapproachesinfinity

you want to find the limit?

68. SolomonZelman

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

69. SolomonZelman

|dw:1435010612413:dw|

70. SolomonZelman

|dw:1435010912898:dw|

71. SolomonZelman

when you multiply something by a number smaller than 1 you make it smaller

72. SolomonZelman

this is why 2/n^2 is greater than n!/n^n

73. SolomonZelman

so call n!/n^n A(n) and 2/n^2 B(n)

74. SolomonZelman

by definition of comparison test, if B(n) is larger and converges, then A(n) will also converges