Geometric
P-Series
Ratio test
Alternating Series
Error Bound
100

Determine whether the geometric series with the first term being 4 and r = 2 converges or diverges. If the series converges, find it sum

|2| ≥ 1, Series Diverges

100

Determine if  Σ∞n=1 1/n2  converges or diverges

since p = 2, and p > 1, the series converges by p-series test

100

determine whether an=1/n! converges or diverges

limn->∞      an+1/an=0 since 0<1, the series converges absolutely. 

100

Determine if the series converges: Σ(-1)n * (1/n)

Yes, it converges conditionally by the Alternating Series Test (Leibniz’s Test).

  • 1/n is decreasing and approaches 0.
  • The series alternates signs.

However, it does not converge absolutely because Σ1/n diverges (harmonic series).

100

estimate the error in approximating ln(2)=Σ(-1)n+1/n using the first 3 terms

1-1/2+1/3=5/6

|error| <= |an+1| =1/4

|ln(2)-5/6|<=.25

200

A geometric series first term is 3, and r = -1/3. Does this series converge? If so, find the sum.

S = 3 / (1- (-1/3)) 

S = 9/4


200

Given, Σ∞n=1  1/n0.8, determine if the series converges or diverges

since p = 0.8,  0 < p ≤ 1, the series diverges by p-series test

200

Find the interval of convergence for Σ(2x)n/n!

limn->oo |(2x)n+1n!/((2x)n(n+1)!)| = limn->oo|2x/(n+1)|=0 -> 0<1 :. converges absolutely

200

Determine whether the series Σ(-1)n+1/n2 converges absolutely or conditionally. 

The series converges absolutely. 

* Ignore the sign alternator so the new series is 1/n2.

* Absolute series Σ1/n2 is a p-series with p=2>1, so it converges.

*therefore, the original alternating series converges absolutely.

200

How many terms of Σ(-1)n+1/n are needed to approximate ln(2) with error less than .001. 

1/(n+1) <.001 --> n+1>1000 --> n>999

at least 1000 terms. 

300

 Consider the geometric series 

Σ∞n=1  5 * (3/2)n-1

Determine if the series converges or diverges


since |r| ≥ 1, r = 3/2, the series diverges

300

Given Σ∞n=1  n/(n3+n), determine if the series converges or diverges

Converges by direct comparison test.

new series is bigger and convergent

300

Find the radius of convergence for Σn2(x-6)n/(3n)!

limn->∞ |(n+1)2(x-6)/(3)(n2)| = limn->∞ |(x-6)/3|

|(x-6)/3| < 1

|x-6| < 3 

R = 3


300

Does the following series converge absolutely, conditionally, or diverge? Σ(-1)n * n/(n+1)

The series diverges. 

  • The terms n/(n+1) --> 1 != 0 as n approaches infinity, so the limit of the terms does not approach zero.
  • Diverges by the nth-term test for divergence.
300

Approximate sin(.5) using its Maclaurin series up to the x3 term, then find an upper bound for the error. 

R3(x)=f(4)(c)/4!*x4 --> f(4)=sin(x), and |sin(X)| <=1: |R3(.5)| <= 1/4!(.5)4 = 1/384 ~.0026

400

A geometric series has an infinite sum of 20 and a first term of 12. Determine the common ratio r and determine if the series converges

S = a/(1-r) , given a = 12 and S = 20

--> 12/(1-r) = 20 --> 12 = 20(1-r) --> 1-r = 3/5

--> r = 2/5

since |r| < 1, r = 2/5, the series converges

400

Given , Σ∞n=1  1/n2x, determine for what values of x, does the series diverge

a series diverges when 0 < x ≤ 1

--> 0 < 2x < 1 --> 0 < x ≤ 1/2

400

For the series ∑ (x-7)n/3n find the endpoints of the interval of convergence

limn->oo|((x-7)n+1/3n+1) / ((x-7)n / 3n)| 

limn->oo|(x-7)/3| = |x-7| < 1

-1< x-7 < 1

6 < x < 8

Endpoints are x=6 and x=8

400

Determine convergence and type for: 

Σ(-1)nln(n)/n

The series converges conditionally.Let an=ln(n)/n:It's positive and decreasing for n>=3. limn->∞  ln(n)/n=0  So the series passes the alternating series test. but absolute convergence fails because Σln(n)/n diverges via integral test. 

400

Estimate error when approximating cos(1) using cos(x) ~ 1-x2/2!+x4/4!

R4(x) = f(5)(c)(x)5/5!, f(5)(x) = sin(x)

|R4(1)|<=1/120~.0083

500

A geometric series  Σ∞n=0   arn has a first term of 10 and converges to an infinite sum of 15. Find the common ratio r and verify the convergence

since he series starts at n=0, first term is 10 

--> a/(1-r) = 15 --> 10/(1-r) = 15 --> 10=15(1-r)

--> 1-r = 2/3 --> r = 1/3

since |r| < 1, r = 1/3, the series converges

500

Given Σ∞n=1  26597834/n.125x + 17, determine the values of p in which the series diverges

a series diverges when 0 < x ≤ 1

--> 0 < .125x + 17 ≤ 1 --> -17 < .125  ≤ -16

--> -17(8)< x ≤ -16(8)

500

Solve for the interval of convergence of the following series: Σoon=1 n2/4n(x+2)n

|an+1/an|=|(n+1)2/4n+1(x+2)n+1*4n/n2(x+2)n| 

|an+1/an|=|(n+1)2(x+2)/4n2|

limn->oo|an+1/an|=limn->oo|(n+1)2(x+2)/4n2|

                        =|(x+2)/4|limn->oo((n+1)2/n2)

                          |(x+2)/4| < 1 

                          |(x+2)| <4 

                         -4 < x+2 < 4 

                            -6 < x < 2 

Therefore, the interval of convergence is (-6,2)

500

Given: Σ(-1)n1/(√n + 1)

Determine if the series converges absolutely, conditionally, or diverges. 

The series converges conditionally. 

an=1/(√n+1) is positive, decreasing, and converges to zero. 

compare to 1/√n, which diverges since p=1/2

thus since the new series is smaller, the original series converges conditionally. 

500

The Maclaurin series for arctan(x)=Σ(-1)nx2n+1/(2n+1) how many terms are required to estimate arctan(.5) with an error < .0001

Alternating series, |error| <= |.52n+3/(2n+3)| < .0001

n=1: error~.00625, n=2: error~.0011, n=3: error~.00022, n=4: error~.000044

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