anonymous
  • anonymous
Help with series.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Bring it
anonymous
  • anonymous
:)
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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anonymous
  • anonymous
http://gyazo.com/c66486c586b2f191cb027a9bf9967b07.png
anonymous
  • anonymous
Not necessarily. A series that is greater than a convergent series could be either divergent or convergent.
anonymous
  • anonymous
How can I then show it?
anonymous
  • anonymous
Did you determine that second series in your question to be convergent?
anonymous
  • anonymous
No it is divergent...
anonymous
  • anonymous
Well then, in that case then yes, the series would be divergent.
anonymous
  • anonymous
Great.
Jhannybean
  • Jhannybean
Have you learned the tests for convergence? such as Comparison Test, or Test of Divergence...etc.
anonymous
  • 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
  • anonymous
yes,@Jhannybean
Jhannybean
  • Jhannybean
What you stated is an alternating series,And I don't see an alternating series in the one you posted :\
Jhannybean
  • Jhannybean
\[\sum_{n=1}^{∞}\frac{ n^3+\ln(n) }{ \sqrt{n^7+n^2} }\ge \frac{1}{2\sqrt{n}}\]
Jhannybean
  • Jhannybean
The first test you should ALWAYS try is test of divergence. Have you tried that yet?
anonymous
  • 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
  • anonymous
Are you thinking on first or sec series I posted?
Jhannybean
  • Jhannybean
Ohh you posted 2... i thought they were related :\
Jhannybean
  • Jhannybean
Let's see...
anonymous
  • 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
  • Jhannybean
Ohh okay,use the alternating series test :)
anonymous
  • anonymous
@FutureMathProfessor Your teacher is a wise man
Jhannybean
  • 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
  • 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
  • anonymous
So teh series are conditional convergence.
anonymous
  • anonymous
*the
Jhannybean
  • Jhannybean
Yeah... if you don't get the same value after taking the absolute value of an, then the series is conditionally convergent.
Jhannybean
  • Jhannybean
If the absolute value and an itself result in the same answer, then the series is absolutely convergent
Jhannybean
  • 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
  • anonymous
Okay.. Thank you @Jhannybean
Jhannybean
  • Jhannybean
No problem...do you understand how to solve it now?
anonymous
  • anonymous
Year it think it make sense. But I will spend some time on alternating series tomorrow.
Jhannybean
  • Jhannybean
ok :)

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