PROBLEM SET X
“FINAL DESTINATION”
DUE FRIDAY, 9 DECEMBER
Exercise 71. Suppose (an )n≥0 and (bn )n≥0 two sequences of positive real numbers. Which of the following is greater?
In each case, give examples when equality obtains, as well as examples when it does not.
lim sup(an + bn ) and
n→∞
lim sup(an bn ) and
n→∞
lim sup an + lim sup bn ;
n→∞
n→∞
lim sup an
lim sup bn ;
n→∞
lim sup an−1
n→∞
−1
lim sup an
and
;
n→∞
n→∞
lim inf(an + bn ) and
n→∞
lim inf(an bn ) and
n→∞
lim inf an−1
and
n→∞
lim inf an + lim inf bn ;
n→∞
n→∞

‹
‹
lim inf an
lim inf bn ;
n→∞
n→∞

‹−1
lim inf an
.
n→∞
Exercise 72. Suppose (an )n≥0 a sequence of positive real numbers. Prove that
Give an example to show that the positivity condition is necessary.
P
an converges only if
Exercise 73. Suppose (an )n≥0 a sequence of positive real numbers. Prove that if
P
P
an2 converges.
an converges, then
1
X
1 + an
diverges, and
an
X
1 + an
converges.
Exercise? 74. Consider the function f : [0, 1]
R given by
(
x −x if x 6= 0;
f (x) :=
1
if x = 0.
Prove that f is continuous, prove that the series
∞
X
m −m
m=1
converges, and show that
Z
1
f (x) =
0
∞
X
m −m .
m=1
Definition. If k is a positive integer, then the generalized harmonic numbers Hn (k) are the partial sums of the
harmonic series:
n 1
X
.
Hn (k) :=
k
j =1 j
1
2
DUE FRIDAY, 9 DECEMBER
Exercise? 75. Consider the sequence (xn )n≥1 with
xn := Hn (1) − log n.
Prove that this sequence converges by writing γ := lim supn→∞ xn and finding a number c > 0 such that |xn − γ | ≤
c/n for any n ≥ 1.
Definition. Suppose k an integer. Then the k-th polylogarithm is the series
∞ 1
X
Lik (z) :=
zn.
k
n
n=1
Exercise 76. Compute the radius of convergence of Lik , and note that it is at least 1.
Exercise 77. Show that if k is a negative integer, then (z − 1)1−k Lik (z) is a polynomial of degree −k with integral
coefficients. What are the roots of this polynomial? What can you say about its coefficients?
Exercise 78. Compute the radius of convergence of the power series
∞
X
M k (z) :=
Hn (k)z n .
n=1
Show, moreover, that within this radius of convergence, one has
(z − 1)M k (z) = Lik (z).