1
DEF 1
A sequence {𝑎𝑎𝑛𝑛 } has the limit L
𝑙𝑙𝑙𝑙𝑙𝑙 𝑎𝑎𝑛𝑛 = 𝐿𝐿
𝑛𝑛→∞
𝑜𝑜𝑜𝑜
𝑎𝑎𝑛𝑛 → 𝐿𝐿
𝑎𝑎𝑎𝑎
𝑛𝑛 → ∞
if we can make the terms 𝑎𝑎𝑛𝑛 as close to 𝐿𝐿 as we like by taking 𝑛𝑛 sufficiently large.
If 𝑙𝑙𝑙𝑙𝑚𝑚 𝑎𝑎𝑛𝑛 exists , we say the sequence is convergent.
𝑛𝑛→∞
Otherwise, we say the sequence is divergent.
THM
If lim |𝑎𝑎𝑛𝑛 | = 0 , then 𝑙𝑙𝑙𝑙𝑙𝑙 𝑎𝑎𝑛𝑛 = 0
𝑥𝑥→∞
𝑛𝑛→∞
This thm is true only if L = 0
DEF
If the sequence {𝑺𝑺𝒏𝒏 } is convergent and lim 𝑆𝑆𝑛𝑛 = 𝑆𝑆 exists as a real number,
𝑛𝑛→∞
then the series {𝒂𝒂𝒏𝒏 } is called convergent and we write
∞
� 𝑎𝑎𝑛𝑛 = 𝑆𝑆
𝑛𝑛=1
▪ S is the sum of the series.
∞
If lim 𝑆𝑆𝑛𝑛 is nonexistent, then the series � 𝑎𝑎𝑛𝑛 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅 and has no sum.
𝑛𝑛→∞
𝑛𝑛=1
2
∞
Summary of Tests for Infinite Series Convergence �� 𝑎𝑎𝑛𝑛
𝑛𝑛=1
𝑜𝑜𝑜𝑜
nth – term test
If 𝑙𝑙𝑙𝑙𝑙𝑙 𝑎𝑎𝑛𝑛 ≠ 0, then the series is divergent.
𝑛𝑛→∞
If 𝑙𝑙𝑙𝑙𝑙𝑙 𝑎𝑎𝑛𝑛 = 0, then the series may converge or diverge, so you need to use different test.
𝑛𝑛→∞
.
Geometric Series Test
∞
If the series has the form � 𝑎𝑎 𝑟𝑟
𝑛𝑛=1
𝑛𝑛−1
∞
or � 𝑎𝑎 𝑟𝑟 𝑛𝑛 ,
𝑛𝑛=0
then the series converges if |𝑟𝑟| < 1 and diverges otherwise
𝑎𝑎
If the series converges, then it converges to
1 − 𝑟𝑟
p – series test (proof by integral test)
If the series has the form of
∞
�
𝑛𝑛=1
1
, then the series converges if p > 1 and diverges otherwise.
𝑛𝑛𝑝𝑝
When 𝑝𝑝 = 1, the series is the divergent Harmonic series.
Integral Test
If 𝑎𝑎𝑛𝑛 = 𝑓𝑓(𝑛𝑛) and 𝑓𝑓(𝑥𝑥) is 𝐜𝐜ontinuous, 𝐩𝐩ositive and 𝐝𝐝ecreasing function on the interval [1, ∞)
∞
∞
then � 𝑎𝑎𝑛𝑛 and improper integral � 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 either both converge or both diverge.
𝑛𝑛=1
If convergent, upper bound is:
∞
∞
1
∞
� 𝑎𝑎𝑛𝑛 ≤ 𝑎𝑎1 + � 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑛𝑛=1
1
and
S = � 𝑎𝑎𝑛𝑛 = 𝑎𝑎1 + 𝑎𝑎2 + ⋯ + 𝑎𝑎𝑛𝑛 + 𝑎𝑎𝑛𝑛+1 + 𝑎𝑎𝑛𝑛+2 + 𝑎𝑎𝑛𝑛+3 + ⋯ = 𝑆𝑆𝑛𝑛 + 𝑅𝑅𝑛𝑛
𝑛𝑛=1
𝑆𝑆𝑛𝑛 is Estimation of the sum and 𝑅𝑅𝑛𝑛 is the Remainder
∞
∞
� 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 ≤ 𝑅𝑅𝑛𝑛 ≤ � 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑
𝑛𝑛+1
𝑛𝑛
⟹
∞
∞
𝑆𝑆𝑛𝑛 + � 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 ≤ 𝑆𝑆 ≤ 𝑆𝑆𝑛𝑛 + � 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑
𝑛𝑛+1
𝑛𝑛
∞
� 𝑎𝑎𝑛𝑛 �
𝑛𝑛=0
3
Direct Comparison Test (DCT)
∞
∞
▪ If 𝑎𝑎𝑛𝑛 ≤ 𝑏𝑏𝑛𝑛 and � 𝑏𝑏𝑛𝑛 converges, then � 𝑎𝑎𝑛𝑛 converges.
𝑛𝑛=1
𝑛𝑛=1
∞
∞
▪ If 𝑎𝑎𝑛𝑛 ≥ 𝑏𝑏𝑛𝑛 and � 𝑏𝑏𝑛𝑛 diverges, then � 𝑎𝑎𝑛𝑛 diverges.
𝑛𝑛=1
𝑛𝑛
1
Ex: compare 𝑛𝑛 to 𝑛𝑛 ,
2
2
𝑛𝑛=1
1
1
𝑛𝑛2
𝑛𝑛
to
,
to 2
(𝑛𝑛 + 3)2
𝑛𝑛3 + 1
𝑛𝑛3 (𝑛𝑛2 + 3)2
Limit Comparison Test (LCT)
(may be used instead of DCT most of the time)
If 𝑎𝑎𝑛𝑛 , 𝑏𝑏𝑛𝑛 > 0 and lim �
𝑛𝑛→∞
𝑎𝑎𝑛𝑛
𝑏𝑏𝑛𝑛
� 𝐨𝐨𝐨𝐨 lim � � equal any finite number,
𝑛𝑛→∞ 𝑎𝑎𝑛𝑛
𝑏𝑏𝑛𝑛
then either both � 𝑎𝑎𝑛𝑛 and � 𝑏𝑏𝑛𝑛
converge or diverge.
Use this test when you cannot compare term by term because the rule of sequence is “too UGLY” but you can still
find a known series to compare with it.
1
3𝑛𝑛2 + 2𝑛𝑛 − 1
to 3 (you can disregard the leading coefficient and all nonleading terms,
Ex: compare
5
4𝑛𝑛 − 6𝑛𝑛 + 7
𝑛𝑛
1
𝑛𝑛2
looking only at the condensed degree of the leading terms: 5 = 3
𝑛𝑛
𝑛𝑛
Ratio Test
If lim �
𝑛𝑛→∞
𝑎𝑎𝑛𝑛+1
� = L (real number or ∞) , then
𝑎𝑎𝑛𝑛
(𝑖𝑖) � 𝑎𝑎𝑛𝑛 converges absolutely (and hence converges) if L < 1
(𝑖𝑖𝑖𝑖) � 𝑎𝑎𝑛𝑛 diverges if L > 1 (or ∞)
(𝑖𝑖𝑖𝑖𝑖𝑖) the test is inclonclusive if L = 1 (use another test)
Use this test for series whose terms converge rapidly, for instance those involving exponentials and/or factorials!!!
Root Test
If
𝑛𝑛
lim �|𝑎𝑎𝑛𝑛 | = L (real number or ∞) , then
𝑛𝑛→∞
(𝑖𝑖) � 𝑎𝑎𝑛𝑛 converges absolutely (and hence converges) if L < 1
(𝑖𝑖𝑖𝑖) � 𝑎𝑎𝑛𝑛 diverges if L > 1 (or ∞)
(𝑖𝑖𝑖𝑖𝑖𝑖) the test is inclonclusive if L = 1 (use another test)
𝑒𝑒 2𝑛𝑛
Use this test for series involving n𝑡𝑡ℎ powers. Ex) � 𝑛𝑛
𝑛𝑛
4
Alternating series test
∞
∞
𝑛𝑛=1
𝑛𝑛=1
If the series has the form � 𝑎𝑎𝑛𝑛 = �(−1)𝑛𝑛−1 𝑏𝑏𝑛𝑛
𝑖𝑖)
𝑖𝑖𝑖𝑖)
0 ≤ 𝑏𝑏𝑛𝑛+1 ≤ 𝑏𝑏𝑛𝑛
𝑙𝑙𝑙𝑙𝑙𝑙 𝑏𝑏𝑛𝑛 = 0
𝑛𝑛→∞
𝑏𝑏𝑛𝑛 > 0, then the series converges if
𝑓𝑓𝑓𝑓𝑓𝑓 𝑎𝑎𝑎𝑎𝑎𝑎 𝑛𝑛 ∈ 𝑍𝑍 + (decreasing terms)
If either of these conditions fails, the test fails, and you need use a different test.
If convergent, a partial sum Sn is used as an approximation to the total sum S.
𝑅𝑅𝑛𝑛 is the Remainder: |𝑅𝑅𝑛𝑛 | = |𝑆𝑆 − 𝑆𝑆𝑛𝑛 | ≤ 𝑏𝑏𝑛𝑛+1 , and the sum 𝑆𝑆 lies: 𝑆𝑆𝑛𝑛 − 𝑏𝑏𝑛𝑛+1 ≤ 𝑆𝑆 ≤ 𝑆𝑆𝑛𝑛 + 𝑏𝑏𝑛𝑛+1
ABSOLUTE AND CONDITIONAL CONVERGENCE (for LOCO series +, - ,+,+,+,-,-,-)
∞
∞
𝑛𝑛=1
𝑛𝑛=1
If �|𝑎𝑎𝑛𝑛 | converges, then � 𝑎𝑎𝑛𝑛 is 𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪.
∞
If � 𝑎𝑎𝑛𝑛 is absolutly convergent then it is (just) convergent. The converse of this statement is false.
𝑛𝑛=1
∞
∞
∞
𝑛𝑛=1
𝑛𝑛=1
𝑛𝑛=1
If �|𝑎𝑎𝑛𝑛 | diverges but � 𝑎𝑎𝑛𝑛 converges , then � 𝑎𝑎𝑛𝑛 is 𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂 𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐠𝐠𝐠𝐠𝐠𝐠𝐠𝐠
∞
�
𝑛𝑛=1
(−1)𝑛𝑛−1
is conditially convergent. Harmonic series is divergent, alternating harmonic series is convergent.
𝑛𝑛
Remember, if you are asked to find the ACTUAL sum of an infinite series, it must either be a Geometric
𝑎𝑎
series ( S =
) or a Telescoping Series (requires expanding and canceling terms).
1−𝑟𝑟
The only other tests that allows us to approximate the infinite sum are the Integral test and the Alternate Series
th
Test. We can find the n partial sum Sn for any series.