Questions Asked

privetek:
interpret the following statement: points have a location, but no size or direction; nonzero vectors have a size and direction, but no locat…
 updated one year ago
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privetek:
use comparison test to determine whether the series converges:
sum(k=1>infinity) [ ( (3k)^(6/7) )/(sqrt(k)+1)^4 ]
 updated one year ago
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privetek:
use comparison test to determine whether the series converges:
sum(k=1>infinity) [ ( (3k)^(6/7) )/(sqrt(k)+1)^4 ]
 updated one year ago
 1 reply
 No Medals Yet

privetek:
use comparison test to determine whether the series converges:
sum(k=1>infinity) [ ( (3k)^(6/7) )/(sqrt(k)+1)^4 ]
 updated one year ago
 No Medals Yet

privetek:
determine whether the series converges:
sum(k=1 > infinity) [1/(1+lnk)]
 updated one year ago
 14 replies
 1 Medal

privetek:
determine whether the series converges:
sum(k=1 > infinity) [1/(1+lnk)]
 updated one year ago
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privetek:
f(x) = e^x, a=0; approximate e^0.5 using the linear approximating polynomial
 updated one year ago
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privetek:
determine convergence or divergence. for converging, state absolute or conditional convergence:
sum(k=1 > infinity) [(1)^(k+1)(10^k)]/(k!…
 updated one year ago
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privetek:
determine absolute or conditional convergence:
sum(k=1 > infinity) [(1)^(k+1)(10^k)]/(k!)
 updated one year ago
 22 replies
 No Medals Yet

privetek:
determine convergence or divergence. for converging series, state absolute or conditional (explain why):
sum(k=2 > infinity) [(1)^k/(k^2…
 updated one year ago
 17 replies
 2 Medals

privetek:
integral of [(7x)/(e^x)] as x goes from 1 to infinity
 updated one year ago
 17 replies
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privetek:
determine whether the series converges or diverges:
 updated one year ago
 12 replies
 1 Medal

privetek:
determine convergence or divergence of the series:
 updated one year ago
 3 replies
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privetek:
use the root test to determine if series converges or diverges:
sum(x=1>infinity) [((n!)^2n)/((2n)!)^n]
 updated one year ago
 3 replies
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privetek:
use integral test to determine whether the series converges:
sum(n=1 > infinity) [1/((ln5)^n)]
 updated one year ago
 3 replies
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privetek:
assume the sequence converges, find its limit:
a sub1=6, a sub(n+1)=72/(1+a subn)
 updated one year ago
 2 replies
 No Medals Yet

privetek:
find the limit of (7n8)/(2sqrtn) as n goes to infinity
 updated one year ago
 13 replies
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privetek:
determine if series converges or diverges using the comparison test:
sum(n=1 to infinity)[3/(8n+7sqrt(n))]
 updated one year ago
 11 replies
 1 Medal

privetek:
limit of (sinn)^3/(3^n) as n goes to infinity
 updated one year ago
 1 reply
 1 Medal

privetek:
what is the limit of (5+(1)^n)/5 as n goes to infinity
 updated one year ago
 8 replies
 1 Medal