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tux
Group Title
Decide convergence / divergence:
sum n=1 to infinity 1/(n*sqrt(n)+2)
 2 years ago
 2 years ago
tux Group Title
Decide convergence / divergence: sum n=1 to infinity 1/(n*sqrt(n)+2)
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.2
converges for sure
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
since exponent in denominator is greater than 1
 2 years ago

tux Group TitleBest ResponseYou've already chosen the best response.0
How to solve using comparison test or integral test?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
integral test is just integrate from 1 to infinity
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
comparison is a limit of the ratio of .... \[\frac{a_{n+1}}{a_n}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
actually that is "ratio test" comparison means compare to something else you know converges (or diverges)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
of is the comparison a b_n of another ..... yeah lol
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
\[\sum_1^{\infty}\frac{1}{n\sqrt{n}+2}\]compare \[\frac{1}{n\sqrt{n}+2}<\frac{1}{n^{\frac{3}{2}}}\] second sum converges
 2 years ago
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