anonymous
  • anonymous
alternating series of (-1)^n (ln(n))/n^2 can i compare the series to a 1/n^2 convergent series and then say that by limit comparison test the series absolutely converges?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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NotTim
  • NotTim
What subject is this?
anonymous
  • anonymous
math...
NotTim
  • NotTim
wow. then I must be super dumb.

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anonymous
  • anonymous
? amm is calculus 2 ... alternating series.. ... .. amm i think im still doing math lol
Zarkon
  • Zarkon
\[\ln(n)\le \sqrt{n}\]
Zarkon
  • Zarkon
\[\frac{\ln(n)}{n^2}\le\frac{\sqrt{n}}{n^2}=\frac{1}{n^{3/2}}\]
experimentX
  • experimentX
Yup, Zarkon is right, 1/n^(3/2) converges so left hand side must also converge. we can show that (-1)^n * (ln(n))/n^2 converges if (ln(n))/n^2 converges, since absolute value is greater than the other alternating value
anonymous
  • anonymous
ok but why compareto to ((n)^1/2))/(n^2)???
Zarkon
  • Zarkon
why not...it works
experimentX
  • experimentX
because ((n)^1/2))/(n^2) is a convergent series and your series is less than it
anonymous
  • anonymous
got it thanks guys...

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