SME
Calculus II · Problem type
Radius and interval of convergence
Use this for “Find the interval of convergence of Σ (x − 1)ⁿ ÷ (4ⁿ√n)” (√n is n0.5).
Diagram
−1 ≤ x < 5: −1 ≤ x and x < 5, both true at once: the numbers between.
The closed circle takes in −1; the open circle leaves out 5.
This is an example. Type your own values to start a new problem.
Σ
(x
−
)ⁿ
ⁿ
n
R
Radius of convergence
worked out (not typed)
L
Left end, c − R
worked out (not typed)
U
Right end, c + R
worked out (not typed)
ρ
Ratio limit at either end (always 1)
worked out (not typed)
at L
Series at the left end
worked out (not typed)
at U
Series at the right end
worked out (not typed)
Formulas
R
=
b
3 = 3
L
=
c
−
R
−1 = 2 − 3
U
=
c
+
R
5 = 2 + 3
ρ
=
|
U
−
c
|
÷
b
1 = |5 − 2| ÷ 3
at
L
=
converges
if
p
>
0,
else
diverges
(ratio
ρ
:
no
verdict)
1 > 0, so converges
at
U
=
converges
if
p
>
1,
else
diverges
(ratio
ρ
:
no
verdict)
1 ≤ 1, so diverges
Assumptions
•
The series is Σ from n = 1 to ∞ of (x − c)ⁿ ÷ (bⁿnᵖ), with b > 0 and p ≥ 0 (√n is p = 0.5).
•
Ratio test: |aₙ₊₁ ÷ aₙ| = (|x − c| ÷ b) × (n ÷ (n + 1))ᵖ, which tends to |x − c| ÷ b. It converges when that is under 1.
•
At the ends the ratio is 1 and the test says nothing: x = c + b gives Σ 1/nᵖ and x = c − b gives Σ (−1)ⁿ/nᵖ.
Step-by-step
WE KNOW
c = 2, b = 3, p = 1
FIND
radius of convergence (R), left end, c − R (L), right end, c + R (U), ratio limit at either end (always 1) (ρ), series at the left end (at L), series at the right end (at U)
Step 1 · Find radius of convergence (R)
Use R = b
The
ratio
of
terms
tends
to
|x
−
c
|
÷
b
.
It
is
under
1
exactly
when
|x
−
c
|
<
b
,
so
the
series
converges
within
b
of
the
center.
R
=
3
Step 2 · Find left end, c − R (L)
Use L = c − R
The
interval
reaches
R
to
the
left
of
the
center.
L
=
2
−
3
L
=
−1
Step 3 · Find right end, c + R (U)
Use U = c + R
And
R
to
the
right.
U
=
2
+
3
U
=
5
Step 4 · Find ratio limit at either end (always 1): ρ
Use ρ = |U − c| ÷ b
At
either
end
|x
−
c
|
=
b
,
so
the
ratio
tends
to
exactly
1
and
the
ratio
test
decides
nothing:
each
end
needs
a
test
of
its
own.
ρ
=
|5
−
2|
÷
3
ρ
=
|3|
÷
3
ρ
=
3
3
ρ
=
1
Step 5 · Find series at the left end (at L)
Use at L = converges if p > 0, else diverges (ratio ρ: no verdict)
At
x
=
c
−
b
,
x −
c
b
=
−1,
leaving
Σ
(−1)ⁿ/nᵖ.
Its
sizes
1/nᵖ
shrink
to
0,
so
the
alternating
series
test
says
it
converges:
the
end
is
in.
1 > 0: converges
at
L
=
converges
if
1
>
0,
else
diverges
(ratio
1:
no
verdict)
at
L
=
converges
Step 6 · Find series at the right end (at U)
Use at U = converges if p > 1, else diverges (ratio ρ: no verdict)
At
x
=
c
+
b
,
x −
c
b
=
1,
leaving
the
p
-series
Σ
1/nᵖ.
With
p
≤
1
it
diverges:
the
end
is
out.
1 ≤ 1: diverges
at
U
=
converges
if
1
>
1,
else
diverges
(ratio
1:
no
verdict)
Interval of convergence: −1 ≤ x < 5
at
U
=
diverges
Check
Put the numbers back in. Both sides should match.
3 = 3 ✓
−1 = 2 − 3 ✓
5 = 2 + 3 ✓
1 = |5 − 2| ÷ 3 ✓
1 > 0, so converges ✓
1 ≤ 1, so diverges ✓
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