Question

For what values of p is the following series convergent?

Answer 1

I like these type of questions as I don't know how to solve them. :P

Answer 2

lol

Answer 3

\[\sum_{n=1}^{\infty} \frac{ (-1)^{n-1} }{ n ^{p}}\]

Answer 4

it's hard to tell but it's n to the p power

Answer 5

But you have told us simply.. :P

Answer 6

Hints :
1) alternating harmonic series
2) p-series

Answer 7

i got p>0 before during class but totally forgot how to replicate :/

Answer 8

Well, if \(p \le 1\) then the series will diverge, \(p \gt 1\) series converges,

Answer 9

series converges when p = 1

Answer 10

\(p=0\) for this one, it'd be a harmonic series

Answer 11

if* right?

Answer 12

I thought it diverges if p = 1

Answer 13

convergent only when p>1

Answer 14

That's what I thought too.

Answer 15

in this case it converges because of the (-1)^n-1

Answer 16

Lmao.

Answer 17

p between (0, 1) is interesting

Answer 18

1 -1/2 + 1/3 -1/4 + 1/5 ... this converges

1 + 1/2 + 1/3 + ... this diverges

Answer 19

it converges for 0 < p < 1 , by alternating series test , i think , since
sequence |an| is a decreasing sequence

Answer 20

Yes p>0 works!

Answer 21

the series converges conditionally for 0 < p<=1 , and converges absolutely for p > 1

Answer 22

alternating series is sufficinet i think, we can forget about p-series completely

Answer 23

right

Answer 24

Agree

Answer 25

p-series is only required if we want to talk about absolute convergence

Answer 26

Btw, use this for convergent series tests. http://jacobi.math.wvu.edu/~hjlai/Teaching/Math156_Website/Series%20Cheat%20Sheet.pdf

Answer 27

Bro, that's sick, I'm totally saving that haha.

Answer 28

That's sikh bruv.

Answer 29

Haha. That's neat though, thanks :D

Answer 30

hey

Answer 31

that was like compressing half of calcII in one page !

Answer 32

no problem, I printed one out for myself for Calc 2. Once you understand the forms, you understand what the heck the function is doing

Answer 33

\[\sum_{n=3}^{\infty} \frac{ n+2 }{ (n+1)^3 }\] this would require p series right?

Answer 34

Only if you could reduce it to me 1/n^p

Answer 35

be*

Answer 36

Use a comparison test I think, you should get 1>0 then \[\sum_{n=3}^{\infty} \frac{ 1 }{ n^2 }\]?

Answer 37

This one is pretty awesome too. And colorful http://www.math.hawaii.edu/~ralph/Classes/242/SeriesConvTests.pdf

Answer 38

And it has a whole bunch of related problems too, haha. I solved some of these I remember! :)

Answer 39

I like the previous cheat sheet, this one seems to complicated haha.

Answer 40

\[\sum_{n=3}^{\infty} \frac{ n+2 }{ (n+1)^3 } =\sum_{n=3}^{\infty} \frac{ 1 }{ (n+1)^2 } + \dfrac{1}{(n+1)^3} \]

Answer 41

shift the index and use p-series direct w/o comparison

Answer 42

How did you reduce that lol

Answer 43

when you have n+2 ontop and n+1 on bottom?

Answer 44

n+2 = (n+1)+1

Answer 45

Oh =_=

Answer 46

That's smart haha, thanks.

Sorry @alexwee123

Answer 47

yeah thats pretty clever, also you can use limit comparison test , that might be faster

Answer 48

(n+2) / (n+1)^3
behaves like
n/ n^3 for large n , or 1/n^2

so you can limit comparison with that

Answer 49

Yup, that's what I meant to say by comparison test :)