PHZ 3113 Fall 2010
Homework #1, Due Friday, September 4
α n
. What is lim an ? What is lim an ?
1. Let an = 1 +
n→∞
n
n→0
As in many cases like this, it is convenient to look at the logarithm: if the log has a limit,
then so does the original, and the log of the limit is the limit of the log.
For small n,
h α
α
n i
n ln 1 +
= n ln
+ ln 1 +
= n(ln α − ln n) + O(n2 ) → 0.
n
n
α
For large n,
Thus,
α 1
α
n ln 1 +
=n
−
n
n 2
α n
lim 1 +
= 1,
n
n→0
α2
1 α2
→ α.
+
·
·
·
=
α
−
2 n
n2
α n
lim 1 +
= eα .
n→∞
n
2. Decide whether the following sums converge or diverge. Present the reasons for your
choices.
(a)
∞
X
n=2
1
n ln n ln(ln n)
The derivative of ln(ln(ln x)) is
d
1
d
1
1 d
1
ln(ln(ln x)) =
ln(ln x) =
ln x =
.
dx
ln(ln x) dx
ln(ln x) ln x dx
ln(ln x) ln(x) x
Thus, this sum diverges, although very slowly, by the integral test,
N
X
n=2
1
∼
n ln n ln(ln n)
Z N
dx
= ln(ln(ln N)) → ∞.
x ln x ln(ln x)
The divergence is very slow: The sum of 10 terms is 2.53945; the sum of 100 terms is 3.11987;
1000 terms is 3.35467; 106 terms is 3.66112 (see plot).
(b) This sum in fact does converge,
∞
X
sin n
π−1
=
,
n
2
n=1
but none of the tests we have studied is really sufficient to prove this. Since | sin n| ≤ 1, it
P
is bounded by |an | ≤ bn = 1/n, but
bn does not converge (that harmonic series again).
The factor sin n introduces fluctuating signs, but they do not alternate systematically. The
integral converges,
Z ∞
sin x
π
dx = ,
x
2
0
and that is probably the best justification.
Writing sin n as a sum,
∞
∞
∞
∞
∞
X
X
sin n X 1 X (−1)k
(−1)k X 2k
2k+1
=
n
=
n
n
n
(2k + 1)!
(2k + 1)!
n=1
n=1
k=0
k=0
n=1
does not lead anywhere useful that I can see. (Although it does lead Mathematica to conclude
that the value of the sum is − 12 , since the sum on n is ζ(−2k) [ζ is the Riemann zeta function];
and ζ(0) = − 12 , ζ(−2k) = 0 for integer k > 0. The failure is exchanging the order of sums,
which is not justified here.)
(c)
∞ X
n=1
1
1
−
2n + 1 2n
This is more or less the alternating harmonic series, which we already know converges (yes,
it is OK to cite results obtained in class),
∞ X
n=1
Or, combine terms,
1 1 1 1 1 1 1
1
−
= − + − + − + · · · = ln 2 − 1.
2n + 1 2n
3 2 5 4 7 6
∞ X
n=1
X
1
1
1
−
,
=−
2n + 1 2n
2n(2n + 1)
which converges by the integral test, for one.
∞
n=1
3. Consider the following alternating series
2
1
2
1
2
1
2
1
2
1− √ + √ − √ + √ − √ + √ − √ + √ − √ +··· .
2
3
4
5
6
7
8
9
10
Does the series converge?
√
This series alternates in sign, and the general term goes to zero as 1/ n, but it does not do
so monotonically in amplitude, and in fact the sum diverges. To see this, write the series as
∞
∞
X
(−1)n+1 X
1
√
√
2
;
−
n
2n
+
1
n=1
n=1
√
the first term is a converging alternating series [it sums to 2( 2 − 1)ζ( 12 ) = 1.2098 . . .], but
the second term diverges, by say the integral test.
4. A quantum rotator with moment of inertia I has energy levels ` = `(` + 1) h̄2 /2I, with
degeneracy 2`+1, where ` is an integer and h̄ is the (reduced) Planck constant. The partition
function for a system with these energy levels is
Z=
∞
X
(2` + 1)e−β` ,
`=0
where β = 1/kT .
(a) Determine for what values of β the series converges.
Let α = βh̄2 /2I; then the sum is
Z=
∞
X
(2` + 1)e−α`(`+1).
`=0
The ratio of successive terms is
(2` + 3)e−α(`+1)(`+2)
(2` + 1)e−α`(`+1)
=
1 + 3/2` −2α(`+1)
e
→ e−2α(`+1),
1 + 1/2`
which vanishes exponentially for large `; by the ratio test the series converges for any finite
value of α, and so any finite value of β or T .
(b) Compute an approximate value for Z when T is large. Compute an approximate value
for Z when T is small.
When T is large, or β is small, convergence does not begin until ` is large. In this case, it is
sufficient to approximate the sum by the integral,
Z
Z1 = d` (2` + 1) e−α`(`+1).
Let u = `(` + 1), du = (2` + 1) d`, and
Z ∞
1
2IkT
Z1 =
du e−αu = =
.
α
h̄2
0
When T is small, β is large, the magnitude of terms falls off rapidly, and only the first few
are necessary:
Z0 = 1 + 3 e−2α.
The solid line shows the full function Z(α); the short-dashed line shows Z0 , valid for small
argument, and the long-dashed line shows Z1 , valid for large argument.
Z(α)
5
4
3
2
1
0.0
0.5
1.0
1.5
α
2.0
2.5
3.0
P
5. Take the alternating harmonic series (−1)n+1 /n and sum it in the following order:
take three odd (positive) terms, followed by the first even (negative) term, followed by three
more odd terms, one more even term, three odd terms, one even term, etc.
s=1+
1 1 1 1 1
1
1
1
1
1
1
+ − + + +
− +
+
+
− +··· .
3 5 2 7 9 11 4 13 15 17 6
Does this series converge? If so, what is the value of the sum? Since infinite precision, even
if achievable, is not practical, make some choices and include an estimate of the accuracy of
your result.
Write the sum as
X
∞ ∞
X
1
1
1
1
15 − 92n + 108n2
s=
+
+
−
.
=
6n − 5 6n − 3 6n − 1 2n
2n(6n − 1)(6n − 3)(6n − 5)
n=1
n=1
For large n, the sum of the four-term pattern goes as 108n2 /[2n(6n)3 ] = 1/4n2 , which
converges by the integral test or comparison test or ratio comparison test. The integral
provides an estimate of the remainder as
Z ∞
dn
4
= .
2
N
N 4n
Sums of 10, 100, and 1000 terms are s10 = 1.21789, s100 = 1.23996, s1000 = 1.24220;
corrected sums, including the 1/4N estimate of the remainder, are s10 = 1.24289, s100 =
1.24245, s1000 = 1.24245.
The exact answer is 12 ln 12 = 1.24245.