SME
Calculus II · Problem type
The ratio test
Use this for “Does Σ n(0.5)ⁿ converge? Use the ratio test, and find the sum if there is one.”
Chart
Terms: 0.5, 0.5, 0.375, 0.25, 0.1563 (aₙ = n·0.5ⁿ).
a₅ = 5 × 0.5⁵ = 0.1563
a₆ = 6 × 0.5⁶ = 0.09375
a₆
a₅
=
0.09375
0.1563
=
0.6;
the
ratio
tends
to
|r|
=
0.5
<
1:
the
series
converges
S₅ = 1.781
S = 2; S₅ is 0.2188 from it.
This is an example. Type your own values to start a new problem.
Σ
n(
r
)ⁿ
=
S
n
Term number
example
aₙ
Term n
worked out (not typed)
aₙ₊₁
Next term
worked out (not typed)
aₙ₊₁/aₙ
Ratio of next term to term n
calculated
L
Limit of |aₙ₊₁ ÷ aₙ|
worked out (not typed)
C
What the ratio test says
worked out (not typed)
Formulas
aₙ
=
n
×
r
n
0.15625
=
5
×
0.5
5
aₙ₊₁
=
(
n
+
1)
×
r
n
+ 1
0.09375
=
(5
+
1)
×
0.5
5 + 1
aₙ₊₁/aₙ = aₙ₊₁ ÷ aₙ
0.6
=
0.09375
0.15625
aₙ₊₁/aₙ
=
(
n
+
1)
×
r
n
0.6
=
(5
+
1)
×
0.5
5
L
=
|
r
|
0.5 = |0.5|
C
=
converges
if
L
<
1,
diverges
if
L
>
1
0.5 < 1, so converges
S
=
r
÷
(1 −
r
)
2
2
=
0.5
÷
(1 − 0.5)
2
Assumptions
•
The terms are aₙ = n·rⁿ for n = 1, 2, 3, … with r fixed. When r is negative the signs alternate.
•
aₙ₊₁ ÷ aₙ = ((n + 1) ÷ n) × r, and (n + 1) ÷ n tends to 1, so |aₙ₊₁ ÷ aₙ| tends to L = |r|.
•
The ratio test: L < 1 means the series converges, L > 1 means it diverges, and L = 1 decides nothing.
•
When |r| < 1 the sum is r ÷ (1 − r)²: differentiate Σ xⁿ = 1 ÷ (1 − x), then multiply by x.
Step-by-step
WE KNOW
r = 0.5, n = 5
FIND
term n (aₙ), next term (aₙ₊₁), ratio of next term to term n (aₙ₊₁/aₙ), limit of |aₙ₊₁ ÷ aₙ| (L), what the ratio test says (C), sum of the series (S)
Step 1 · Find term n (aₙ)
Use aₙ = n × rⁿ
Put
the
term
number
n
into
the
rule
n
·rⁿ.
aₙ
=
5
×
0.5
5
aₙ = 5 × 0.03125
aₙ = 0.15625
Step 2 · Find next term (aₙ₊₁)
Use aₙ₊₁ = (n + 1) × rⁿ⁺¹
Put
the
next
term
number,
n
+
1,
into
the
same
rule.
aₙ₊₁
=
(5
+
1)
×
0.5
5 + 1
aₙ₊₁
=
6
×
0.5
6
aₙ₊₁ = 6 × 0.015625
aₙ₊₁ = 0.09375
Step 3 · Find ratio of next term to term n (aₙ₊₁/aₙ)
Use aₙ₊₁/aₙ = aₙ₊₁ ÷ aₙ
Divide the next term by this one.
aₙ₊₁/aₙ
=
0.09375
0.15625
aₙ₊₁/aₙ
=
(
n
+
1)
×
r
n
=
6
×
0.5
5
aₙ₊₁/aₙ = 0.6
Step 4 · Find limit of |aₙ₊₁ ÷ aₙ| (L)
Use L = |r|
As
n
grows,
n
+ 1
n
tends
to
1,
so
the
size
of
the
ratio
tends
to
the
size
of
r
.
L
=
|0.5|
L
=
0.5
Step 5 · Find what the ratio test says (C)
Use C = converges if L < 1, diverges if L > 1
With
L
<
1
the
terms
shrink
at
least
as
fast
as
a
geometric
series
with
ratio
under
1,
so
the
sum
is
finite.
0.5 < 1: converges
C
=
converges
if
0.5
<
1,
diverges
if
0.5
>
1
C
=
converges
Step 6 · Find sum of the series (S)
Use S = r ÷ (1 − r)²
Σ
xⁿ
from
n
=
0
is
1
1 − x
.
Its
derivative
is
Σ
n
·xⁿ⁻¹
=
1
÷
(1 − x)
2
,
and
times
x
that
is
Σ
n
·xⁿ.
S
=
0.5
÷
(1 − 0.5)
2
S
=
0.5
0.5
2
S
=
0.5
0.25
S
=
2
Check
Put the numbers back in. Both sides should match.
0.15625
=
5
×
0.5
5
✓
0.09375
=
(5
+
1)
×
0.5
5 + 1
✓
0.6
=
0.09375
0.15625
✓
0.6
=
(5
+
1)
×
0.5
5
✓
0.5 = |0.5| ✓
0.5 < 1, so converges ✓
2
=
0.5
÷
(1 − 0.5)
2
✓
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