SME
Calculus II
Sequences, series and convergence tests
Use this for “Estimate Σ 1/n² with 10 terms. How far off can it be?”
Chart
Terms:
1,
0.25,
0.1111,
0.0625,
…,
0.01
(aₙ
=
1
n
2
).
a₁₀
=
1
10
2
=
0.01
S₁₀ = 1.55
Band: 1.641 ≤ S ≤ 1.65 (S = 1.645 lies in it)
S = 1.645; S₁₀ is 0.09517 from it.
This is an example. Type your own values to start a new problem.
S
N
=
1
+
1
2
+
1
3
+
…
+
1
N
=
SN
C
Converges or diverges
worked out (not typed)
L
Lower bound on the sum
worked out (not typed)
U
Upper bound on the sum
worked out (not typed)
Formulas
SN
=
N
Σ
n
=1
1/
n
p
1.5498
=
10
Σ
n
=1
1/
n
2
C
=
converges
if
p
>
1,
else
diverges
2 > 1, so converges
L
=
SN
+
1
÷
((
p
−
1)
×
(
N
+ 1)
p
− 1
)
1.6406768
=
1.5498
+
1
÷
((2
−
1)
×
(10 + 1)
2 − 1
)
U
=
SN
+
1
(
p
− 1) ×
N
p
− 1
1.6497677
=
1.5498
+
1
(2 − 1) ×
10
2 − 1
Assumptions
•
The terms 1/nᵖ are positive and shrink as n grows, so the integral test applies to Σ 1/nᵖ.
•
The terms left out after N add to more than the integral of x⁻ᵖ from N + 1 to ∞ and less than the one from N to ∞.
•
With p ≤ 1 those integrals have no finite value, so the series diverges and no band exists.
Step-by-step
WE KNOW
p = 2, N = 10
FIND
partial sum of N terms (SN), converges or diverges (C), lower bound on the sum (L), upper bound on the sum (U)
Step 1 · Find partial sum of N terms (SN)
Use SN = Σ from n = 1 to N of 1/nᵖ
Add
the
first
N
terms
one
by
one:
1,
then
1
2ᵖ
,
then
1
3ᵖ
,
and
so
on.
SN
=
10
Σ
n
=1
1/
n
2
SN = 1 + 0.25 + 0.1111 + … + 0.01 = 1.5498
SN = 1.54977
Step 2 · Find converges or diverges (C)
Use C = converges if p > 1, else diverges
With
p
>
1
the
integral
of
x
−
p
from
1
to
∞
is
finite,
so
by
the
integral
test
the
sum
is
finite
too.
2 > 1: converges
C
=
converges
if
2
>
1,
else
diverges
C
=
converges
Step 3 · Find lower bound on the sum (L)
Use L = SN + 1 ÷ ((p − 1) × (N + 1)ᵖ⁻¹)
The
terms
left
out
add
to
more
than
the
integral
of
x
−
p
from
N
+
1
to
∞,
which
is
1
÷
((
p
−
1)
(
N
+ 1)
p
− 1
).
L
=
1.54977
+
1
÷
((2
−
1)
×
(10 + 1)
2 − 1
)
L
=
1.5498
+
1
11
1
L
=
1.5498
+
1
11
L
=
1.5498
+
0.09091
L
=
1.6406768
Step 4 · Find upper bound on the sum (U)
Use U = SN + 1 ÷ ((p − 1) × Nᵖ⁻¹)
The
terms
left
out
add
to
less
than
the
integral
of
x
−
p
from
N
to
∞,
which
is
1
(
p
− 1)
N
p
− 1
.
U
=
1.54977
+
1
(2 − 1) ×
10
2 − 1
U
=
1.5498
+
1
10
1
U
=
1.5498
+
1
10
U
=
1.5498
+
0.1
U
=
1.6497677
Check
Put the numbers back in. Both sides should match.
1.54977
=
10
Σ
n
=1
1/
n
2
✓
2 > 1, so converges ✓
1.6406768
=
1.54977
+
1
÷
((2
−
1)
×
(10 + 1)
2 − 1
)
✓
1.6497677
=
1.54977
+
1
(2 − 1) ×
10
2 − 1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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