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GayWolf420
June, 2025

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For the smart people out there.

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HHB
Posts: 683 - Karma: 14
Im bored doing hw, and i read threads about smart people and calculus. lets see what kinda knowledge we got in the ns forums

so for all you people. ill give you mad karma if you get it right, of course i already know it, i have G status in calc.

just pickin a random problem and random chapter


Series (cant make the symbol for summation so ill use E)

E (2+ (-1)^n)/(n(sqrt(n)) from n=1 to infinity and beyond!!!!!!!111one!!!




yes im a bit of a nerd...
16 years ago
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Alehandro
Posts: 818 - Karma: 12
jesus christ
is that seriously a problem??
16 years ago
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HenniferLopez
Posts: 1319 - Karma: 136
5?
16 years ago
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woolymammoth
Posts: 242 - Karma: 13
let me get back to you in a week, my class is starting series on tuesday
16 years ago
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RudyGarmisch
Posts: 25444 - Karma: 13,856
Get that shit out, I didnt complete college level math courses for more math.
16 years ago
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Elg
Posts: 2998 - Karma: 3,881
I was looking at that problem, and all i noticed was "sqrt" in there. Squirt?
16 years ago
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HHB
Posts: 683 - Karma: 14
ahaha thats square root, not a soda
16 years ago
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zerospinskier
Posts: 4867 - Karma: 350
I haven't taken the series calc in a while, but do you want to know if it converges, or what it converges to?

You could use the alternating series test.
-its monotonic decreasing
-the lim as n goes to infiniti is 0.

So it should converge.

That's all I got. I don't really remember.
16 years ago
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zerospinskier
Posts: 4867 - Karma: 350
Ooooo wait, it's PLUS (-1)^n. Hang on I gotta rethink
16 years ago
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skierdudeguy
Posts: 6574 - Karma: 81
for the record, that doesnt alternate. im not sure if you were trying to make it, but (2+(-1)^n) (or n-1 i dont remember) will either give you 1 or 3... not an alternating series i dont think. but fuck, i got a c in calc 2
16 years ago
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rhcp
Posts: 2598 - Karma: 18
Fuck, I forget how to do this stuff and I just started a math grading job at my university where I grade tests from a calc II class.
16 years ago
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skierdudeguy
Posts: 6574 - Karma: 81
and im pretty sure (that if im right) it doesnt converge
16 years ago
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zerospinskier
Posts: 4867 - Karma: 350
Yeah exactly, that plus sign fucks everything up. I thought it was multiplied at first.
16 years ago
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cclan
Posts: 742 - Karma: 19
to be quite honest something like this doesn't look like it's going to be much use in your everyday life
16 years ago
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HHB
Posts: 683 - Karma: 14
epic. no it doesnt alternate.
im just looking for converge or diverge
pics to show ur work or it didnt happen!!! haha



integral test....
16 years ago
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zerospinskier
Posts: 4867 - Karma: 350
It's just like 1 /  (n)sqrt(n)    which is really easy, but I don't remember how haha.
16 years ago
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lalalalaura
Posts: 6753 - Karma: 68
squirt baby squirt baby squirt
16 years ago
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Static
Posts: 8598 - Karma: 3,972
Click for larger image
16 years ago
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hellomynameisjef
Posts: 5794 - Karma: 167
seven
16 years ago
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HHB
Posts: 683 - Karma: 14

integral test....

16 years ago
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Jwenz
Posts: 8893 - Karma: 7,634
ALright well using the comparison you can just make it 1/n sqrt n which is just n^-3/2

then you integrate from [0, infinity)

Substitute t for infinity

Giving you -2/3t^1/2 + 0

Because the equation with t goes to infinity the series diverges...i think

series was my downfall last semester
16 years ago
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An10a_Eng
Posts: 2505 - Karma: 26
that is correct
16 years ago
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wellpreparedrare
Posts: 651 - Karma: 11
The tops of all equations will either be 1 or 3, seeing that -1 raised to anything above zero will either be -1 or 1; add 2 to either -1 or 1 and you'll get 1 or 3.
The bottom of the equation will be increasing in sequence (from 1 to infiniti) with all equations.
The series will be getting smaller, but always be positive, seeing that eventually 1 or 3 will be divided by infinity-times-its-square-root. 1 or 3 is will be divided into a shitload of little pieces, and the size of one of those pieces is what each equation will equal.
In other words, your equation will eventually reach infiniti, seeing that the series will always be positive and stretches to infiniti.
For all of you that don't know what a series is, all you have to do is add each sum of the equation to the left of the 'E'. Since 'n' is going from 1 to infiniti, you'll get infiniti sums. Hence, the 'series' will be a series of infiniti equation, all in the order from one to infiniti, and each sum being added to the next in the series.
16 years ago
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wellpreparedrare
Posts: 651 - Karma: 11
this only makes sense if you've read the text and gotten the definitions n shit.
16 years ago
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HHB
Posts: 683 - Karma: 14
that is not correct

comparison test shows that

Original integral
16 years ago
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Mac.
Posts: 953 - Karma: 24
We had a problem like that on our AP Challenge today.
16 years ago
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TheQuailman
Posts: 7382 - Karma: 148

I'm only a junior in highschool and I haven't done this yet but...

NUMERATOR

2 + (-1)^n

So sub infiniti for n and your either going to get -1 or 1 depending on whether n is odd or even.  So the numerator is either 1 or 3 (a constant)

DENOMINATOR

n(radical(n))

So sub infiniti for n and it will always increase to infiniti.

So the finale would be..

(1 or 3)/infiniti

which gets closer and closer to zero as infiniti gets larger

So....

Divergent? 

16 years ago
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wellpreparedrare
Posts: 651 - Karma: 11
exactly, but i don't know if divergent is what your teacher wants to see, but that's whatever, the sum is infiniti. if you place 'n' on an x-axis and the sum as n increases along the y-axis, you'll get a graph going up and to the right, which is to positive infiniti.
16 years ago
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Yung_Gnarley
Posts: 9096 - Karma: 3,071
e l e v e n
16 years ago
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.KYLE.
Posts: 1891 - Karma: 59
heres a better question... how could you use that in real life?
16 years ago
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fssnivek
Posts: 283 - Karma: 10
LOL
+10 for that one
does squirt soda still exist?
16 years ago
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