Math 1321
Week 2 Lab Worksheet
Due Thursday 01/24
1. Warmup: Determine whether the following series is convergent. H int: Is it absolutely
convergent?
∞
X
(−1)n n2
2n
n=1
2
Solution:
for absolute convergence by the Ratio
Test
where an = 2nn . Thus
Checking
2
n2 (1+ 1 )2
(n+1)2 2n
= 2n+1 · n2 = 12 (n+1)
= limn→∞ 12 n2 n =
we have an+1
. Therefore limn→∞ an+1
an n2
an 1
2
< 1. By the Ratio Test the series is absolutely convergent and therefore the
alternating series is also convergent.
2. Estimating an Alternating Series: Closed form solutions to series (even convergent
ones) cannot always be explicitly computed. In these cases, it is best to have a bound
on the
error one would achieve by truncating the series at the nth term. That is, if
P∞
s = n=1 an and s cannot be computed explicitly, we would like to know how many
terms n we need, to approximate s to within a certain accuracy. Use the Alternating
Series Estimation Theorem to answer the following questions.
P
(−1)n ln(n+1)
(a) How many terms are required to approximate the series ∞
to within
n=1
n
5 decimal places?
(b) How many terms are required to approximate the series
within 5 decimal places?
P∞
n=1
(−1)n ln(n+1)e−n
n
to
(c) Which series required more terms to approximate accurately? Give a brief explanation as to why.
Solution: (a) The Alternating Series
Theorem states the remainder Rn
P∞Estimation
n
of an alternating series of the form n=1 (−1) bn satisfies |Rn | ≤ bn+1 . For the series
in part (a) bn = ln(n+1)
. Requiring bn < 1 × 10−5 we find (by trial and error or CAS
n
loop) n = 1416361. Thus we need n = 1416360 terms to achieve this accuracy.
−n
(b) Similar to part (a) for bn = ln(n+1)e
< 1 × 10−5 we find n = 11 and therefore
n2
we need 10 terms to achieve this accuracy.
(c) The first series required many more terms to approximate accurately. The reasoning for this lies in the decay of the terms bn . The second series terms decreased
much more rapidly due to the exponential growth of the denominator. Thus the
convergence was much more rapid in the second series.
3. Bonding energy of ionic crystals: The work W (in Joules) required to pull one
Cl− away from one Na+ atom with starting x distance apart is given by W (x) = C/x.
Likewise, to pull one Na+ away from another Na+ is simply a negative work −W (x)1 .
Real 3D Crystals of Na+ and Cl− form a square lattice like that shown below.
1
The work required W (x) can be derived from integrating the effect of moving an atom against the
attracting and repulsing force fields generated between the atom pairs.
Thus to compute the bonding energy resulting from pulling apart all the atoms in the
crystal away from a single Na+ atom, would entail summing the work W (x) from all
the distances x from the central Na+ atom, resulting in an alternating series. Focusing
on the Na+ ion at the center of the cubic unit cell, it is noted that this positive
√ ∗ ion is
−
∗
+
attracted to 6 Cl ions at distance r , repelled
√ ∗ by 12 Na ions at distance 2r away,
then attracted to 8 Cl- ions at distance 3r away, and so on out to the edge of the
crystal.
12
C 6
8
6
24
− √ + √ − √ + √ + ...
EN a + = ∗
r 1
2
3
4
5
C
= ∗ M,
r
where the constant M ≈ 1.7475 is the approximate value of the alternating series, and
is termed the Madelung constant after its discoverer.
(a) Compute the bonding energy of a more simplified (non-real) crystal consisting of a
long chain of alternating Na+ -Cl− atoms, each spaced r∗ apart.
(b) Compute the Talyor series expansion to ln(2) centered at the point a = 1 and
compare with (a).
Solution: (a)
EN a +
C 2 2 2 2
= ∗
− + − + ...
r 1 2 3 4
∞
2C X (−1)n+1
= ∗
r n=1
n
(b) The Taylor series representation of natural log centered at the point a = 1 with
x=2
∞
X (−1)n+1
1
1
2
6
ln(2) = ln(1) + − + − + + . . . =
1 2! 3!
4!
n
n=1
So, the alternating harmonic series converges to ln(2). Thus, the bond energy is
EN a+ = C2rln(2)
∗