Question

Help with series.

Answer 1

Bring it

Answer 2

:)

Answer 3

When
\[\sum_{n=1}^{∞}\frac{ n^3+\ln(n) }{ \sqrt{n^7+n^2} }\ge1/2*n^{-1/2}\]
is the series divergent. Right?

Answer 4

http://gyazo.com/c66486c586b2f191cb027a9bf9967b07.png

Answer 5

Not necessarily. A series that is greater than a convergent series could be either divergent or convergent.

Answer 6

How can I then show it?

Answer 7

Did you determine that second series in your question to be convergent?

Answer 8

No it is divergent...

Answer 9

Well then, in that case then yes, the series would be divergent.

Answer 10

Great.

Answer 11

Have you learned the tests for convergence? such as Comparison Test, or Test of Divergence...etc.

Answer 12

What about this
\[\sum_{n=1}^{∞}\frac{ (-1)^n }{ \ln(n)^{1/2} }\]
have can I show it is divergent or Conditional Convergence? It is not abs. convergent.

Answer 13

yes,@Jhannybean

Answer 14

What you stated is an alternating series,And I don't see an alternating series in the one you posted :\

Answer 15

\[\sum_{n=1}^{∞}\frac{ n^3+\ln(n) }{ \sqrt{n^7+n^2} }\ge \frac{1}{2\sqrt{n}}\]

Answer 16

The first test you should ALWAYS try is test of divergence.

Have you tried that yet?

Answer 17

Want to know what my differential equations teacher said about series?
"There should be a separate study for series, as it most certainly is not mathematics"

Answer 18

Are you thinking on first or sec series I posted?

Answer 19

Ohh you posted 2... i thought they were related :\

Answer 20

Let's see...

Answer 21

I think we solve the first one, it war divergent.. Now I am asking for help for this one. :)
\[\sum_{n=1}^{∞}\frac{ (-1)^n }{ \ln(n)^{1/2} }\]
It is a alternating series right?

Answer 22

Ohh okay,use the alternating series test :)

Answer 23

@FutureMathProfessor
Your teacher is a wise man

Answer 24

so \[\large a_{n} = \frac{1}{\sqrt{1n(n)}} \] now is this function increasingor decreasing? you find that out by taking its derivative.

Then take the limit, \[\large \lim_{n \rightarrow \infty}\frac{1}{\sqrt{\ln(n)}} \]\[\large \lim_{n \rightarrow \infty} \frac{1}{\sqrt{\ln(\infty)}} = \frac{1}{\infty} = 0=C\]

Answer 25

Alternating Series Test \[\large \sum_{n=1}^{\infty}(-1)^n\cdot a_{n} \] means that if \(\large a_{n} \) is decreasing and \(\large \lim_{n \rightarrow \infty} a_{n} = 0\) then the series converges.

Answer 26

So teh series are conditional convergence.

Answer 27

*the

Answer 28

Yeah... if you don't get the same value after taking the absolute value of an, then the series is conditionally convergent.

Answer 29

If the absolute value and an itself result in the same answer, then the series is absolutely convergent

Answer 30

Use the alternating harmonic series for example. \[\large \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} =C\] but if we take it's absolute value... \[\large \sum_{n=1}^{\infty}|\frac{(-1)^{n-1}}{n} |=D\] because it's a harmonic series.

Answer 31

Okay.. Thank you @Jhannybean

Answer 32

No problem...do you understand how to solve it now?

Answer 33

Year it think it make sense. But I will spend some time on alternating series tomorrow.

Answer 34

ok :)