SME
Calculus II · Problem type
The alternating series error bound
Use this for “Add the first 9 terms of 1 − 1/2 + 1/3 − 1/4 + …. How far from the sum can that be?”
Chart
Terms: 1, −0.5, 0.3333, −0.25, …, 0.1111 (aₙ = (−1)ⁿ⁺¹ ÷ n).
a₉
=
1
9
=
0.1111
a₁₀
=
−(
1
10
)
=
−0.1
S₉ = 0.7456
Error: |S − S₉| ≤ |a₁₀| = 0.1
Band: 0.6456 ≤ S ≤ 0.8456 (S = 0.6931 lies in it)
S = 0.6931; S₉ is 0.05249 from it.
This is an example. Type your own values to start a new problem.
S
N
=
1
−
1
2
+
1
3
−
1
4
+
…
=
SN
N
Terms added
example
b
Error bound (the next term)
worked out (not typed)
L
Lowest the sum can be
worked out (not typed)
U
Highest the sum can be
worked out (not typed)
S
Sum of the series
worked out (not typed)
E
Actual error |S − SN|
worked out (not typed)
Formulas
SN
=
N
Σ
n
=1
(−1)
n
+1
/
n
p
0.7456
=
9
Σ
n
=1
(−1)
n
+1
/
n
1
b
=
1
÷
(
N
+ 1)
p
0.1
=
1
÷
(9 + 1)
1
L
=
SN
−
b
0.6456 = 0.7456 − 0.1
U
=
SN
+
b
0.8456 = 0.7456 + 0.1
S
=
∞
Σ
n
=1
(−1)
n
+1
/
n
p
0.6931
=
∞
Σ
n
=1
(−1)
n
+1
/
n
1
E
=
|
S
−
SN|
0.05249 = |0.6931 − 0.7456|
Assumptions
•
The terms are (−1)ⁿ⁺¹/nᵖ with p more than 0: their sizes 1/nᵖ shrink to 0 and the signs take turns, so the series converges.
•
The partial sums zig-zag about the sum S, so SN is off by at most the size of the first term left out, b = 1/(N + 1)ᵖ.
•
The sum is ln 2 for p = 1, π² ÷ 12 for p = 2 and 7π⁴ ÷ 720 for p = 4. For other p the chart draws it.
Step-by-step
WE KNOW
p = 1, N = 9
FIND
partial sum of N terms (SN), error bound (the next term) (b), lowest the sum can be (L), highest the sum can be (U), sum of the series (S), actual error |S − SN| (E)
Step 1 · Find partial sum of N terms (SN)
Use SN = Σ from n = 1 to N of (−1)ⁿ⁺¹/nᵖ
Add
the
first
N
terms
one
by
one,
the
signs
taking
turns:
1,
then
minus
1
2ᵖ
,
then
plus
1
3ᵖ
,
and
so
on.
SN
=
9
Σ
n
=1
(−1)
n
+1
/
n
1
SN = 1 − 0.5 + 0.3333 + … + 0.1111 = 0.7456
SN = 0.745635
Step 2 · Find error bound (the next term): b
Use b = 1 ÷ (N + 1)ᵖ
Each new term overshoots the sum and the next one comes back, so the sum is never farther from SN than the first term left out.
b
=
1
÷
(9 + 1)
1
b
=
1
10
1
b
=
1
10
b
=
0.1
Step 3 · Find lowest the sum can be (L)
Use L = SN − b
The
sum
is
at
most
b
below
the
partial
sum.
L
=
0.745635
−
0.1
L
=
0.6456
Step 4 · Find highest the sum can be (U)
Use U = SN + b
The
sum
is
at
most
b
above
the
partial
sum.
U
=
0.745635
+
0.1
U
=
0.8456
Step 5 · Find sum of the series (S)
Use S = Σ from n = 1 to ∞ of (−1)ⁿ⁺¹/nᵖ
For
p
=
1
the
series
is
the
Maclaurin
series
of
ln(1
+
x)
at
x
=
1.
S
=
ln(2)
S
=
0.693147
Step 6 · Find actual error |S − SN| (E)
Use E = |S − SN|
How
far
the
partial
sum
really
is
from
the
sum.
It
is
under
the
bound
b
,
as
the
zig-zag
promises.
E
=
|0.693147
−
0.745635|
E
=
|−0.05249|
E
=
0.05249
Check
Put the numbers back in. Both sides should match.
0.745635
=
9
Σ
n
=1
(−1)
n
+1
/
n
1
✓
0.1
=
1
÷
(9 + 1)
1
✓
0.6456 = 0.745635 − 0.1 ✓
0.8456 = 0.745635 + 0.1 ✓
0.693147
=
∞
Σ
n
=1
(−1)
n
+1
/
n
1
✓
0.05249 = |0.693147 − 0.745635| ✓
Topic check
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