tux
  • tux
Decide convergence / divergence: sum n=1 to infinity 1/(n*sqrt(n)+2)
Mathematics
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tux
  • tux
Decide convergence / divergence: sum n=1 to infinity 1/(n*sqrt(n)+2)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
converges for sure
anonymous
  • anonymous
since exponent in denominator is greater than 1
tux
  • tux
How to solve using comparison test or integral test?

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amistre64
  • amistre64
integral test is just integrate from 1 to infinity
amistre64
  • amistre64
comparison is a limit of the ratio of .... \[\frac{a_{n+1}}{a_n}\]
anonymous
  • anonymous
actually that is "ratio test" comparison means compare to something else you know converges (or diverges)
amistre64
  • amistre64
of is the comparison a b_n of another ..... yeah lol
anonymous
  • anonymous
\[\sum_1^{\infty}\frac{1}{n\sqrt{n}+2}\]compare \[\frac{1}{n\sqrt{n}+2}<\frac{1}{n^{\frac{3}{2}}}\] second sum converges

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