World Series
True or False: 1 - 1/2 + 1/3 - 1/4 + 1/5 + ... is a series
True or False: -1, 2, -3, 4, -5, ... is a series
True or False: If
lim_(n->oo) a_n = 0
, then
sum_(n=1)^oo a_n
is convergent.
True, because we are adding the terms, it is a series.
False, because we are only listing the terms, it is a sequence.
False, because taking the limit of
a_n
is the Divergence Test and if that limit = 0, the test is inconclusive and you'll need to try something else!
In order to determine whether a series converges or diverges, what are the three conditions needed to use the Integral Test? How do you check these conditions?
1. Continuous: check that
f(x)
is defined for all values of x we care about (ex: seeing that the denominator never equals 0)
2. Positive: check that
f(x)
is never negative for all values of x we care about
3. Decreasing: find
f'(x)
and show that
f'(x)<0
for all values of x we care about, OR show that
a_(n+1) \leq a_n
is true for all values of n we care about
What is the difference between
the Direct Comparison Test
and
the Limit Comparison Test?
State both.
Suppose that
sum a_n
and
sum b_n
are series with positive terms.
Direct Comparison Test:
(1) If
sum b_n
is convergent and
a_n leq b_n
for all n, then
sum a_n
is also convergent.
(2) If
sum b_n
is divergent and
a_n geq b_n
for all n, then
sum a_n
is also divergent.
Limit Comparison Test:
If
lim_(n->oo) (a_n)/b_n = c
where
c>0
and is a finite number, then either both series converge or diverge.
What is the Alternating Series Test?
If the alternating series
sum_(n=1)^oo (-1)^(n-1)b_n = b_1 - b_2 + b_3 - b_4 + ...
satisfies (1)
b_n
is decreasing for all n and (2)
lim_(n->oo) b_n = 0
then the series is convergent.
What is the Ratio Test? What is the Root Test?
Ratio test:
lim_(n->oo) |(a_(n+1))/a_n | = L
If L < 1, absolutely convergent
If L > 1, divergent
If L = 1, inconclusive
lim_(n->oo) root(n)(|a_n|) = L
(Same L cases as Ratio)
Determine whether the following series converges or diverges (state what test you used).
sum_(n=1)^oo (2n)/(3n+1)
Since
lim_(n->oo) (2n)/(3n+1) = 2/3 != 0
the series diverges by the Divergence Test.
Determine whether the following series is convergent or divergent:
sum_(n=1)^oo 3/n^(5/3)
Convergent p-series, p=5/3 > 1
Determine whether the series converges or diverges:
sum_(n=1)^oo (9^n)/(3+10^n)
Notice that
9^n/(3+10^n) < 9^n/10^n = (9/10)^n
for all
n geq 1
Since
sum_(n=1)^oo (9/10)^n
is a convergent geometric series,
sum_(n=1)^oo 9^n/(3+10^n)
converges by the Direct Comparison Test.
Test the series for convergence or divergence:
2/3-2/5+2/7-2/9+2/11-...
First note that the series can be written as
sum_(n=1)^oo (-1)^(n+1) 2/(2n+1).
Then notice that
b_n = 2/(2n+1) > 0
and
f'(x) = -4/(2x+1)^2 <0
so
{b_n}
is decreasing and
lim_(n->oo) 2/(2n+1) = 0
Then by the Alternating Series Test, the series converges.
Determine if the series is convergent or divergent:
sum_(n=2)^oo ((-1)^(n-1))/(ln n)^n
Ratio Test:
lim_(n->oo) root(n)((1/ln n)^n) = 0
Since L<1, the series is absolutely convergent, so it converges.
Determine whether the series is convergent or divergent. If it is convergent, find its sum.
sum_(n=1)^oo 3^(n+1) 4^(-n)
Since the series can be rewritten as
3 sum_(n=1)^oo (3/4)^n
we notice that it is geometric, and since |r|<1, the series converges to
(9/4)/(1-3/4) = 9
Use the Integral Test (listing the 3 conditions) to determine whether the series is convergent or divergent:
sum_(n=1)^oo 2/(5n-1)
The function
f(x)=2/(5x-1)
is continuous, positive and decreasing on
[1,oo)
so we can use the Integral Test.
Since the improper integral is divergent, the series is also divergent by the Integral Test.
Determine whether the series converges or diverges:
sum_(n=1)^oo 1/(sqrt(n^2+1))
Note that we can compare the series with
sum_(n=1)^oo 1/n
so, using the Limit Comparison Test, we have that
lim_(n->oo) (1/sqrt(n^2+1))/(1/n) = 1 > 0.
Since the harmonic series diverges, so does
sum_(n=1)^oo 1/sqrt(n^2+1)
Test the series for convergence or divergence:
sum_(n=1)^oo (-1)^(n+1) n^2/(n^3+4)
Since
n^2/(n^3+4) > 0
for
n geq 1
and
n^2/(n^3+4)
is decreasing since
f'(x) = (x(8-x^3))/(x^3+4)^2
for
x>2
. Note also that
lim_(n->oo) (n^2)/(n^3+4) = 0
. Thus the series converges by the Alternating Series Test.
Determine if the series converges or diverges
sum_(n=1)^oo (10^n)/((n+1)4^(2n+1))
Ratio Test:
lim_(n->oo) (5n+5)/(8n+6) = 5/8
Since L<1, the series absolutely converges, so it always converges.
Determine whether the following series is convergent or divergent. If convergent, find its sum:
sum_(n=1)^oo (2^(n))/6^(n-1)
After rewriting the series like the following:
6 sum_(n=1)^oo (2/6)^n
we notice that it is geometric and that because |r| > 1, the series diverges.
Determine whether the series is convergent or divergent:
sum_(n=2)^oo 1/(n ln(n))
Because
f(x)
is continuous, positive and decreasing on
[2,oo)
we can use the Integral Test.
Since the improper integral diverges, the series must also diverge by the Integral Test.
Determine whether the series converges or diverges:
sum_(n=1)^oo (n+1)/(n^3+n)
Using the Limit Comparison Test, we note that we can compare our series with
1/n^2
. Since we know that
lim_(n->oo) (n^2(n+1))/(n(n^2+1)) = 1 > 0
and
sum_(n=1)^oo 1/n^2
is a convergent p-series, the series
sum_(n=1)^oo (n+1)/(n^3+n)
also converges.
Test the series for convergence or divergence:
sum_(n=1)^oo (-1)^(n+1) e^(2/n)
Notice that
lim_(n->oo) e^(2/n) = e^0 = 1
, so
lim_(n->oo) (-1)^(n-1)e^(2/n)
does not exist. Therefore
sum_(n=1)^oo (-1)^(n-1)e^(2/n)
diverges by the Divergence Test.
Determine whether the series converges or diverges:
sum_(n=1)^oo (n^10)/((-10)^(n+1))
Ratio Test:
lim_(n->oo) 1/10 (1+1/n)^n = 1/10
Since L<1, the series absolutely converges, so it always converges.
Determine whether the series is convergent or divergent by expressing
sum_(n=1)^oo 3/(n(n+3))
as a telescoping sum. If it is convergent, find its sum.
The series converges and its sum is
11/6
Determine whether the following series is convergent or divergent:
sum_(n=1)^oo (sqrt(n)+4)/n^2
Since the series can be rewritten as
sum_(n=1)^oo (sqrt(n)+4)/n^2 = sum_(n=1)^oo 1/(n^(3/2)) + 4sum_(n=1)^oo 1/n^2
we know that both series are convergent p-series. Therefore the sum of two convergent series must also converge.
Determine whether the series converges or diverges:
sum_(n=1)^oo (e^n+1)/(n e^n+1)
The series diverges - both DCT and LCT work here by comparing to harmonic series
sum_(n=1)^oo 1/n
Test the series for convergence or divergence:
sum_(n=1)^oo (-1)^(n-1) arctan(n)
Since
lim_(n->oo) arctan(n) = pi/2
, so
lim_(n->oo) (-1)^(n-1) arctan(n)
does not exist. Therefore the series
sum_(n=1)^oo (-1)^(n-1) arctan(n)
diverges by the Divergence Test.
Determine if the series converges or diverges
sum_(n=1)^oo (npi^n)/((-3)^(n-1))
Ratio Test:
lim_(n->oo) pi/3 (1+1/n) = pi/3
Since L>1, the series diverges.