Calculus-Sec-9-4-Solution
9.4 Power Series by SGLee - HSKim -JHLee
1-10. Determine the radius of convergence 
and interval of convergence 
of the following series.
1. 
as 
.
Using the Ratio Test, the given series is absolutely convergent
and therefore convergent when 
, and divergent when 
.
If 
, then the series becomes 
, which is divergent.
If 
, then the series becomes 
, which converges by the Alternating Series Test.
Thus, the given power series converges for 
. So, 
and 
.
2. 
as 
Using the Ratio Test, the given series is absolutely convergent
and therefore convergent when 
, and divergent when 
.
If 
, then the series becomes 
, which converges by the Integral Test.
If 
, then the series becomes 
, which converges by the Alternating Series Test.
Thus, the given power series converges for 
. Hence, 
and 
.
4. 
as 
.
Using the Ratio Test, the given series is absolutely convergent
and therefore convergent when 
, and divergent when 
.
If 
, then the series becomes 
, which is divergent.
If 
, then the series becomes 
, which is not convergent.
Thus, the given power series converges for 
. So, 
and 
.
5. 
as 
.
Using the Ratio Test, the given series is absolutely convergent
and therefore convergent when 
, and divergent when 
.
If 
, then the series becomes 
, which converges by the Alternating Series Test.
If 
, then the series becomes 
the negative series, which is divergent.
Thus, the given power series converges for 
. So, 
and 
.
6. 
as 
.
Using the Ratio Test, the given series is absolutely convergent
and therefore convergent when 
, and divergent when 
.
If 
, then the series becomes 
. Since 
, by the Comparison test,
is convergent.
If 
, then the series becomes 
, which converges by the Alternating Sereis Test.
Thus, the given power series converges for 
. So, 
and 
.
7. 
By the Root test,
for all 
.
Thus, the radius of convergence is 
and the interval of convergence is 
.
9. 
as 
.
Using the Ratio Test, the given series is absolutely convergent
and therefore convergent when 
, and divergent when 
.
If 
, then the series becomes 
.
Since 
,
by the Comparison test, 
is divergent.
If 
, then the series becomes 
.
Since 
, 
.
Thus, the given power series converges for 
.
So, 
and 
.
10. 
as 
.
Then, the given power series converges for 
.
So, 
and 
.
11-13. Determine the interval of convergence of a power series representation for the function 
.
11. 
Since this is a geometric series, it converges when 
, that is 
.
Therefore, the interval of convergence is 
.
12. 
Since this is a geometric series, it converges when 
.
Therefore, the interval of convergence is 
.
13. 
.
Since this is a geometric series, it converges when 
.
Therefore, the interval of convergence is 
.
14. Express the function as the sum of a power series by first using partial fractions.
Find the interval of convergence : 
.
Since this is a geometric series, it converges when 
and 
, respectively.
Therefore, the interval of convergence is 
.
15-16. Find a power series representation for the function and determine the radius of convergence.
15. 
The derivative of 
is 
. We have
for 
.
Thus,
We put 
in this equation to determine the value of 
. That is, 
or 
. Thus,
Here 
since the radius of convergence is the same as for the original series.
16. 
The derivative of 
is 
. Now, we have
, for 
.
Thus,
We put 
in this equation to determine the value of 
. Then, 
or 
. Thus,
Here 
since the radius of convergence is the same as for the original series.
17. (a) Find a power series representation for 
. What is the radius of convergence?
(b) Use part (a) to find a power series for 
.
(a) By Example 8 in Section 9.2,
.
(b) 
.
The power series for 
is 
.
By part (a),
.
Thus,
18-19. Evaluate the indefinite integral as a power series and find the radius of convergence.
18. 
.
Thus, the power series is
.
The above power series converges when 
.
Therefore, the interval of convergence is 
.
19. 
Since 
.
Hence,
.
Therefore,
.
3. 
8. 
