Help with series.

- anonymous

Help with series.

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- anonymous

Bring it

- anonymous

:)

- anonymous

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

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## More answers

- anonymous

http://gyazo.com/c66486c586b2f191cb027a9bf9967b07.png

- anonymous

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

- anonymous

How can I then show it?

- anonymous

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

- anonymous

No it is divergent...

- anonymous

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

- anonymous

Great.

- Jhannybean

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

- anonymous

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.

- anonymous

yes,@Jhannybean

- Jhannybean

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

- Jhannybean

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

- Jhannybean

The first test you should ALWAYS try is test of divergence.
Have you tried that yet?

- anonymous

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"

- anonymous

Are you thinking on first or sec series I posted?

- Jhannybean

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

- Jhannybean

Let's see...

- anonymous

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?

- Jhannybean

Ohh okay,use the alternating series test :)

- anonymous

@FutureMathProfessor
Your teacher is a wise man

- Jhannybean

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

- Jhannybean

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.

- anonymous

So teh series are conditional convergence.

- anonymous

*the

- Jhannybean

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

- Jhannybean

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

- Jhannybean

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.

- anonymous

Okay.. Thank you @Jhannybean

- Jhannybean

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

- anonymous

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

- Jhannybean

ok :)

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