For the smart people out there.

HHB

HHB

Member
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...
 
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.
 
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
 
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.
 
epic. no it doesnt alternate.im just looking for converge or divergepics to show ur work or it didnt happen!!! haha

integral test....
 
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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
 
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.
 
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?
 
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.
 
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