Question

Why does a P series converge if P is between 1 and zero??? doesn't it diverge but slowly?

Answer 1

(assuming from n=1 to infinity)

Answer 2

Hmmmmm it doesn't converge if the power is between 0 and 1.

If you take for example the harmonic series, 1/n, it diverges, and that is where p=1.
So in order for a P series to converge it needs to converge faster than the harmonic series (must have a P greater than 1).

Hopefully I'm remembering that correctly :) lol

Answer 3

Yes my books says that it will diverge but this makes no sense to me

why does it have to be faster than the harmonic series? if I start doing 1/n^(1/2) for example, denominator gets bigger and bigger. why doesn't that mean that it will eventually converge?

Answer 4

\[\frac{ 1 }{ \sqrt n }>\frac{ 1 }{ n }\]
\[\frac{ 1 }{ \sqrt2 }>\frac{ 1 }{ 2 }\]
Hmm I think the denominators are getting smaller actually.
Making the actually terms get larger, meaning it doesn't converge quickly enough.

Answer 5

\[\sum_{n=1}^{infinity} \frac{ 1 }{ n^\frac{ 1 }{ 2 } }\]

Answer 6

n=1 : 1
n=2 : .707
n=3: .577
n=4: .5

Answer 7

I think that only happens for the first few terms though, we want to know what happens long term.

1/10 = .1
1/sqrt10 = .316

Answer 8

Hmm yah that is strange though :)

Answer 9

so you agree that it seems like it should converge?

Answer 10

i know that the answer is no but it annoys me that I don't know why

Answer 11

If you were adding the first 7 or 8 terms, it would be less than 1/n, yes.
But if you go any higher than that, the terms become much much larger than 1/n

Answer 12

Did you see the example i gave with n=10? :O

Answer 13

so the point is not if it is getting smaller but if it is larger or smaller than the harmonic at the corresponding term?

Answer 14

You want to eventually be adding 0 with each term, and the size of each term that will get you there fast enough can be determined by the harmonic series.

When p=1, 1/n grows too quickly to converge, the terms don't get small enough fast enough.

When p=1/2, 1/n grows even FASTER than 1/n. Each term is larger than 1/n, so the series converses EVEN SLOWER than 1/n. Meaning it doesn't converge in the long run.

For some reason I was getting confused by your decimals when you put the first few terms out there :) I looked at them wrong. For some reason I was thinking that the 4th term caught up to the 4th term in 1/n but it didn't.

Answer 15

The terms need to be decreasing FASTER than 1 + 1/2 + 1/3 + 1/4 + ... + 1/n.

In the case of the sqareroot, they're decreasing SLOWER than those terms.

Still confused? D':

Answer 16

See how each new term you add is too large?