Section 10.7
Absolute Convergence
Definition of Absolute
verses
Conditional Convergence.
If
If
𝑎𝑛 converges then we say that
𝑎𝑛 converges absolutely.
𝑎𝑛 converges, but 𝑎𝑛 does not converge, then we say that
𝑎𝑛 converges conditionally.
(−1)𝑛
𝑛
For example:
converges conditionally.
It needs the “condition” of alternating to converge.
(−1)𝑛
2𝑛
For example:
converges absolutely.
It does NOT need the “condition” of alternating to converge.
∞
Ex: #22 (read the directions)
𝑛=1
1
𝑛2 + 1
What do you think?
By comparison to the p-series where 𝑛 = 2, this series converges.
It does not alternate, so
|𝑎𝑛 | =
𝑎𝑛 .
Therefore, the series converges absolutely.
∞
Ex: #28
𝑛=1
(−1)𝑛 𝑛!
𝑛𝑛
Do the terms go to zero?
1
𝑛! 𝑛 𝑛 − 1 𝑛 − 2 ∙∙∙ 3 ∙ 2 ∙ 1
<
=
𝑛
𝑛
𝑛
𝑛∙ 𝑛 ∙ 𝑛∙ ∙ ∙𝑛∙𝑛∙𝑛
1
𝑛!
𝑆𝑖𝑛𝑐𝑒 → 0, 𝑠𝑜 𝑑𝑜𝑒𝑠 𝑛
𝑛
𝑛
This series converges
absolutely by the
Ratio Test.
This series converges by the A.S.T.
…but would it converge without the (−1)𝑛 ?
𝑛 + 1 ! 𝑛𝑛
𝑛 + 1 ! (𝑛 + 1)𝑛+1
= lim
lim
𝑛
𝑛→∞ 𝑛! (𝑛 + 1)𝑛+1
𝑛→∞
𝑛! 𝑛
1
1
𝑛 + 1 𝑛! 𝑛𝑛
= <1
=
lim
= lim
1
𝑒
𝑛→∞
𝑛→∞ 𝑛! 𝑛 + 1 𝑛 (𝑛 + 1)
(1 + )𝑛
𝑛
∞
Ex: #32
𝑛=1
(−1)𝑛 23𝑛
7𝑛
∞
=
𝑛=1
(−1)𝑛 8𝑛
7𝑛
8𝑛
lim 𝑛 ≠ 0
𝑛→∞ 7
This series diverges by the nth Term Test.
…and the crowd goes wild!
∞
Ex: #36
𝑛=1
(𝑛!)2
2𝑛 !
Can’t fool me…this series
is NOT alternating!
𝑛 + 1 ! 𝑛 + 1 ! 2𝑛 !
( 𝑛 + 1 !)2 2 𝑛 + 1 !
= lim
lim
2
𝑛→∞
𝑛! 𝑛! 2𝑛 + 2 !
𝑛→∞
𝑛! / 2𝑛 !
𝑛+1 𝑛+1
𝑛 + 1 𝑛! 𝑛 + 1 𝑛! 2𝑛 !
= lim
= lim
𝑛→∞ 2 𝑛 + 1 2𝑛 + 1
𝑛→∞ 𝑛! 𝑛! 2𝑛 + 2 2𝑛 + 1 (2𝑛)!
𝑛+1
1
= lim
= <1
𝑛→∞ 2 2𝑛 + 1
4
This series converges absolutely by the Ratio Test.
∞
Ex: #42
𝑛=1
lim
𝑛𝑛
𝑛→∞ 3𝑛
2
(−1)𝑛+1 𝑛𝑛
Do the terms go to zero?
2
3𝑛
Beats me…
=
Let’s go straight to the Root Test.
lim
𝑛→∞
𝑛
𝑛𝑛
2
3𝑛
= lim
𝑛→∞
𝑛𝑛
3𝑛
2
1/𝑛
𝑛
= lim 𝑛
𝑛→∞ 3
1
=0
= lim
𝑛
𝑛→∞ (ln 3) 3
This series converges absolutely by the Root Test.
∞
Ex: #46 (read the directions)
𝑛=1
7
𝑛=1
(−1)𝑛+1
𝑛𝑛
(−1)𝑛+1
1 1
1
1
1
1
=1− +
−
+
−
+
𝑛
𝑛
4 27 256 3125 46656 823543
≈ .7834305678
The error in an alternating series is less than the next term.
1
1
Namely: 𝑎8 = 8 =
≈ .0000000596046
8
16777216
So our answer is correct
to the 7th decimal place.
𝑆 ≈ .7834306
∞
Ex: #52 (read the directions)
cos 1 =
𝑛=0
(−1)𝑛
2𝑛 !
Five decimal places means the error must be less than .000005
𝑛
2𝑛 !
2
2
3
6
⋮
⋮
8 40,320
9 362,880
8
cos 1 ≈
𝑛=0
1
Need
< .000005
2𝑛 !
… 𝑜𝑟 … 2𝑛 ! > 200,000
𝑎9 is a small enough error.
(−1)𝑛
1 1 1 1
1
1
1
1
=1− + − + −
+
−
+
2𝑛 !
2 4! 6! 8! 10! 12! 14! 16!
≈ .5403023059 rounded to 5 decimal places ≈ .54030
cos 1 ≈ .5403023059, so we actually have
much greater accuracy than was requested.