Math 105/206 - Quiz 6, Apr 10 2015
IMPORTANT: Write your name AND student number somewhere on this sheet.
No calculators, books or notes. Please show your work to get full marks. (10 + 2 marks total)
Problem 1
For each of the following series, say if it converges or diverges and motivate your answer (3 marks each).
∞
X
k=37
∞
X
k2
1
+ 2k + 1
(−1)k
k=1
ek
1
+1
Solution The first series is
∞
X
∞
∞
X
X
1
1
1
=
=
2
2
k + 2k + 1 k=37 (k + 1)
k2
k=37
k=38
and being a p-series with p = 2, it is convergent. Alternatively you can use comparison or limit comparison with
1
. The ratio test won’t work because the limit of the ratios will be 1, and you can’t say anything in that case.
k2
For the second one, it is alternating and the function f (x) = ex1+1 is decreasing and converges to 0 as
x → +∞, so the alternating series test applies. Alternatively, you can see that the series is absolutely convergent,
for example using the ratio test.
Problem 2
Determine the radius of convergence of the following power series centred in 0. (3 marks)
∞
X
(−10)k
k=1
k 3k
· xk
Solution Using the ratio test we find
ak+1 10k+1
k 3k
=
·
ak (k + 1)3(k+1) 10k · |x|
3k
1
k
=
·
· 10|x|
(k + 1)3
k+1
3k
k
and note that k+1
converges to e−3 for k → ∞ (use L’Hôpital after taking the logarithm). So since the first
fraction tends to 0 the limit is
0 · e−3 · 10|x| = 0
for any x. So the radius of convergence is ∞.
Problem 3
Find the Maclaurin series (=Taylor series centred in a = 0) of the function
f (x) =
x
− arctan(x).
1 + 2x2
What is its interval of convergence? (3 marks)
Recall that
∞
X
1
=
xk
1 − x k=0
and
arctan(x) =
∞
X
(−1)k
k=0
x2k+1
2k + 1
for |x| < 1.
Solution
From the formula for the sum of a geometric series we get that
∞
∞
∞
X
X
X
x
2 k
k k 2k
=x
(−2x ) = x
(−1) 2 x =
(−1)k 2k x2k+1
2
1 + 2x
k=0
k=0
k=0
if |2x2 | < 1, and that
arctan(x) =
∞
X
k=0
(−1)k
x2k+1
2k + 1
if |x| < 1, so
∞
∞
∞
2k+1
k
X
X
X
x
k x
k k 2k+1
k 2 (2k + 1) − 1
2
(−1)
(−1)
2
x
−
(−1)
−
arctan(x
)
=
=
· x2k+1 .
1 + 2x2
2k
+
1
2k
+
1
k=0
k=0
k=0
P
k
n
k 2 (2k+1)−1
(note that this means, if you write this series as ∞
n=0 cn · x , that cn = 0 if n is even, and cn = (−1)
2k+1
if n = 2k + 1 is odd)
The interval of convergence is the same as the smallest one for the Taylor series expansions that we used,
that is |2x2 | < 1, which means |x| < √12 , for the geometric series, and |x| < 1 for the arctangent. The smallest
one of these is |x| < √12 , and this is the interval of convergence of our series.