Find the values of x for which the series converges (a) absolutely (b) con-
ditionally P∞n=1(cosechn)x^n
Answers
Step-by-step explanation:
When trying to determine the radius and/or interval of convergence of power series such as these, it is best to use the Ratio Test, which tells us for a series
∑
a
n
, we let
L
=
lim
n
→
∞
∣
∣
∣
a
n
+
1
a
n
∣
∣
∣
.
If
L
<
1
the series is absolutely convergent (and hence convergent)
If
L
>
1
, the series diverges.
If
L
=
1
,
the Ratio Test is inconclusive.
For Power Series, however, three cases are possible
a. The power series converges for all real numbers; its interval of convergence is
(
−
∞
,
∞
)
b. The power series converges for some number
x
=
a
;
its radius of convergence is zero.
c. The most frequent case, the power series converges for
|
x
−
a
|
<
R
with an interval of convergence of
a
−
R
<
x
<
a
+
R
where we must test the endpoints to see what happens with them.
So, here,
a
n
=
(
2
x
−
3
)
n
a
n
+
1
=
(
2
x
−
3
)
n
+
1
=
(
2
x
−
3
)
(
2
x
−
3
)
n
So, apply the Ratio Test:
lim
n
→
∞
∣
∣
∣
∣
⎛
⎝
(
2
x
−
3
)
n
(
2
x
−
3
)
(
2
x
−
3
)
n
⎞
⎠
∣
∣
∣
∣
|
2
x
−
3
|
lim
n
→
∞
1
=
|
2
x
−
3
|
So, if
|
2
x
−
3
|
<
1
, the series converges. But we need this in the form
|
x
−
a
|
<
R
:
∣
∣
∣
2
(
x
−
3
2
)
∣
∣
∣
<
1
2
∣
∣
∣
x
−
3
2
∣
∣
∣
<
1
∣
∣
∣
x
−
3
2
∣
∣
∣
<
1
2
results in convergence. The radius of convergence is
R
=
1
2
.
Now, let's determine the interval:
−
1
2
<
x
−
3
2
<
1
2
−
1
2
+
3
2
<
x
<
1
2
+
3
2
1
<
x
<
2
We need to plug
x
=
1
,
x
=
2
into the original series to see if we have convergence or divergence at these endpoints.
x
=
1
:
∞
∑
n
=
0
(
2
(
1
)
−
3
)
n
=
∞
∑
n
=
0
(
−
1
)
n
diverges, the summand has no limit and certainly doesn't go to zero, it just alternates signs.
x
=
2
:
∞
∑
n
=
0
(
4
−
3
)
n
=
∞
∑
n
=
0
1
diverges as well by the Divergence Test,
lim
n
→
∞
a
n
=
lim
n
→
∞
1
=
1
≠
0
Therefore, the series converges for
1
<
x
<
2