SERIES AND CONVERGENCE TESTS
Introduction
A series is the sum of terms of a sequence,
n
u
r 1
r
.
An infinite series has an infinite number of terms. It is the sum of terms of an infinite sequence,

u
n 1
n
.
If a series is convergent it has a finite sum.
A series that does not have a finite sum is said to be divergent.
Last year we looked at some infinite geometric series that converged (where
𝑛
a
Sn = ∑ uk =
1−r
𝑘=1
for -1<r<1)
Partial Sums
If Sn = ∑𝑛𝑘=1 uk is the nth partial sum and if 𝑆 = lim 𝑆𝑛 exists then the series has a finite sum, S,
𝑛→∞
and is convergent.
If the sum, S, does not exist then the series is divergent.
Conditions for Convergence
For any convergent series

u
n 1
n
, lim 𝑢𝑛 must be zero. Why is this?
n→∞
This condition is necessary but not sufficient. What does this mean?
Example: Harmonic Series

1
n
n 1
Let
1
Sn = ∑nk=1 k
1 1
1
1
1
1 1
1
𝑆2𝑛 − 𝑆𝑛 = (1 + + + ⋯ + +
+ ⋯ + ) − (1 + + + ⋯ + )
2 3
𝑛 𝑛+1
2𝑛
2 3
𝑛
1
1
1
=
+
+ ⋯+
𝑛+1 𝑛+2
2𝑛
1
1
1
≥
+
+ ⋯+
2𝑛 2𝑛
2𝑛
1
= 𝑛( )
2𝑛
1
=
2
1
If lim 𝑆𝑛 = 𝑆 then lim 𝑆2𝑛 = 𝑆 so lim (𝑆2𝑛 − 𝑆𝑛 ) = 0 which is impossible since 𝑆2𝑛 − 𝑆𝑛 ≥ 2.
n→∞
n→∞
CONVERGENCE TESTS
n→∞
One of our aims in studying series is to investigate whether or not a series converges. As noted
above in order for a series to converge lim un must equal 0 but this is a necessary not sufficient
n 
condition.
In order to know whether a series converges we must use a different test. There are four main
convergence test for series that we will study in this course. Read about each one in your text
and as you read make notes on the general features of each test. There is no general recipe to
follow to know which test to use when, but with practice you will start to get the hang of it.
Remember, always check to see if the necessary condition is met first. If it isn’t you do not need
to test anything since you know the series is divergent.
SIMPLE COMPARISON TEST
(NOTE: We have already used this test in the past. It works the same way as the test we used
when dealing with sequences)
∞
Let ∑∞
𝑘=1 𝑎𝑘 and ∑𝑘=1 𝑏𝑘 be series with nonnegative terms and suppose that 𝑎1 ≤ 𝑏1 , 𝑎2 ≤ 𝑏2 ,
𝑎3 ≤ 𝑏3 , …, 𝑎𝑘 ≤ 𝑏𝑘 , …
a) If ∑ 𝑏𝑘 converges, then ∑ 𝑎𝑘 also converges.
b) If ∑ 𝑎𝑘 diverges, then ∑ 𝑏𝑘 also diverges.
Two Informal Principles


Constant terms in the denominator of 𝑢𝑘 can usually be deleted without affecting
convergence or divergence of the series.
If a polynomial k appears as a factor in the numerator or denominator of 𝑢𝑘 , all but the
leading term of the polynomial can usually be discarded without affecting convergence or
divergence of the series.
Ex. Use the comparison test to show that ∑∞
𝑛=1
2𝑛+1 1 𝑛
𝑛
1
(2) is convergent.
Ex. Use the comparison test to show that ∑∞
𝑛=1 2𝑛−1 is divergent.
LIMIT COMPARISON TEST
∞
Let ∑∞
𝑘=1 𝑎𝑘 and ∑𝑘=1 𝑏𝑘 be series with positive terms and suppose that lim
𝑎𝑘
𝑘→∞ 𝑏𝑘
and 𝐿 > 0, then the series both converge or both diverge.
= 𝐿. If L is finite
Ex. Use the limit comparison test to show whether the series converges or diverges.
𝑛+1
a) ∑∞
𝑛=1 𝑛3 +1
𝑛+1
b) ∑∞
𝑛=1 2𝑛2 +1
RATIO TEST
DO NOT WORRY ABOUT THE TELESCOPING OR PARTIAL FRACTIONS PART OF THE EXAMPLES
YET.
Let ∑∞
𝑘=1 𝑢𝑘 be a series with positive terms and suppose that lim
𝑘→∞
𝑢𝑘+1
𝑢
= 𝐿.
a) If 𝐿 < 1 the series converges.
b) If 𝐿 > 1 the series diverges.
c) If 𝐿 = 1 the series may converge or diverge and you must try a different test.
Ex. Use the ratio test to determine convergence or divergence.
1) ∑∞
𝑛=1
2) ∑∞
𝑛=1
3𝑛+1
5𝑛
1
√𝑛
2)
(𝑛!)2
∑∞
𝑛=1 (2𝑛)!
ROOT TEST
𝑘
Let ∑∞
𝑘=1 𝑢𝑘 be a series with positive terms and suppose that lim √𝑢𝑘 = 𝐿.
𝑘→∞
a) If 𝐿 < 1 the series converges.
b) If 𝐿 > 1 the series diverges.
c) If 𝐿 = 1 the series may converge or diverge and you must try a different test.
𝑛
Ex. Use the root test to determine convergence or divergence. Note: lim √𝑛 = 1
𝑛→∞
𝑛
1) ∑∞
𝑛=1 3𝑛
2)
∑∞
𝑛=1
3𝑛
𝑛2 2𝑛
HW: p.664 Exercise Set 10.5 #1-21 odds, 25, 31, 35, 39, 43