Geometric and Nth Term
Integral and P-Series
Comparison Tests
Alternating Series
Ratio and Root
100
Determine whether the series converges or diverges Σ(1+1/n)^n
What is Diverge Limit as n approaches infinity is e e does not equal 0
100
Determine whether the series converges or diverges 1/n^(1/3)
What is Diverge 1/3 < 1
100
Determine whether the series converges or diverges Σ1/((n^3+1)^(1/2))
What is Diverge Use LCT to find that bn = 1/n limit of series as n approaches infinity x 1/bn is positive and finite bn is a harmonic series
100
Determine whether the series converges or diverges. if converges, determine absolutely or conditionally Σ(-1)^n/(2n+1)
What is converges conditionally lim as n approaches infinity = 0 use integral test on 1/(2n+1) u = 2n+1 du = 2 integral diveges
100
Determine whether the series converges or diverges Σn^7/7^n
What is converge use ratio test (n+1)^7/7^(n+1) x (7^n)/n^7 lim (n+1)^7/7n^7 = 0 0<1
200
Determine whether the following series converge or diverge Σ5^n/3^(n+2)
What is Diverge 5^n/3^n+9 r=5/3 5/3>1
200
Determine whether series diverges or converges 2/2(2^(1/2))+2/3(3^(1/2))+2/4(4^(1/2))
What is Converge 2/n(n^(1/2)) n x n^(1/2) = n^(3/2) 3/2>1
200
Determine whether the series converges or diverges. if converges, determine absolutely or conditionally Σ(-1)^n x ln(n)/n
What is converges conditionally lim as n approaches infinity = 0 use integral test u = ln(n) du = 1/n series diverges
200
Determine whether the series converges or diverges Σ(3n/n+2)^n
What is diverges use root test lim as n approaches infinity of 3n/n+2 =3 3>1
300
Determine whether the series converges or diverges Σn^(1/3)/(n+1)^(1/3)
What is diverge Use nth term to get value of 1
300
Determine whether series diverges or converges Σ3n^2 e^-n^3
What is converge u = -n^3 du = -3n^2 lim as n approaches infinity = 1/e^8 1/e^8 < 1
300
Determine whether the series converges or diverges Σ(3n^2+5n^4)(n+1)^-5
What is Diverges use LCT bn = 1/n lim is positive and finite bn is harmonic series
300
Determine whether the series converges or diverges. if converges, determine absolutely or conditionally Σ(-1)^(n+1)(n!)2^3n/(2n)!
What is converges absolutely lim as n approaches infinity = 0 use ratio test to get 8(n+1)/(2n+1)(2n+2) lim = 0 0 < 1
300
Determine whether the series converges or diverges Σ-n^n/n^3n
What is converges use root test -n/n^3 -1/n^2 lim as n approaches infinity is 0 0<1
400
Determine whether the series converges or diverges Σ8n^3-6n^5/12n^4+9n^5
What is diverges use nth term lim = -2/3 2/3 < 1
400
Write out the first three terms of the series and use the integral test to determine whether the series converges or diverges. Σ1/(n(sqrt(ln(n)))
What is 1/(2(sqrt(ln(2))), 1/(3(sqrt(ln(3))), 1/(4(sqrt(ln(4))) + ...; diverges u=ln(n) du=1/n integral from 2 to infinity = infinity
400
Determine whether the series converges or diverges Σ1/n(n^2+1)^(1/2)
What is converge bn = 1/ n^2 lim is positive and finite use p-series for bn 2>1
400
Determine whether the series converges or diverges. if converges, determine absolutely or conditionally Σ(-1)^n e^-n^2
What is converges absolutely lim as n approaches infinity = 0 use dct 1/e^n^2<1/e^n Bigger converges, smaller converges
400
Determine whether the series converges or diverges. Σ(-1)^(n+1) n^2 5^(n+1)/7^n
What is converges use alternating series lim as n approaches infinity is 0 use ratio test lim as n approaches infinity equals 5/7 5/7 < 1
500
Determine whether the series converges or diverges Σ3+4^(n-1)/3^(2n)
What is converges 3+4^(n-1)/9^n 3/9^n+4^(n-1)/9^n 3(1/9)^n + 1/4(4/9)^n both converge because both 1/9 and 4/9 is less than 1
500
find that value of p that makes the series converge 1/(x*(lnx)^p
What is x>1 use integral test to find that series coverges when x >1 u = ln x du = 1/x
500
Determine whether the series converges or diverges Σ1+sin(n!)/n^2
What is converge 1/n^2 + sin(n!)/n^2 1/n^2 converges by p-series sin(n!)/n^2 converges by DCT or ratio
500
let f be a positive, continuous, and decreasing function for x is greater than and equal to 1, such that an = f(n). it the series an coverges to S, then the remainder Rn = S - Sn is bounded by 0 < Rn < integral of f(x) from N to infinity. Find N such that Rn < 0.001 for the convergent series. Σ1/n^2+1
What is N greater than or equal to 1004 Rn is less than or equal to integral of 1/x^2+1 from N to infinity integral is arctanx from N to infinity artan(infinity) - (arctanN) pi/2 - artanN <0.001 arctan > 1.569
500
Determine whether the series converges or diverges. Σn^2/(2n-1)!
What is converges lim (n+1)^2/(2(n+1)-1)! x (2n-1)!/n^2 lim (n+1)^2/(2n+1)(2n)(n^2) = 0 0<1
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