Physics 451
Fall 2004 Homework Assignment #6 — Solutions
Textbook problems: Ch. 5: 5.2.6, 5.2.8, 5.2.9, 5.2.19, 5.3.1 Chapter 5 5.2.6 Test for convergence ∞ X a) (ln n)−1 n=2
As in all these convergence tests, it is good to first have a general idea of whether we expect this to converge or not, and then find an appropriate test to confirm our hunch. For this one, we can imagine that ln n grows very slowly, so that its inverse goes to zero very slowly — too slowly, in fact, to converge. To prove this, we can perform a simple comparison test. Since ln n < n for n ≥ 2, we see that an = (ln n)−1 > n−1 since the harmonic series diverges, and each term is larger than the corresponding harmonic series term, this series must diverge. Note that in this and all subsequent tests, there may be more than one way to prove convergence/divergence. Your solution may be different than that given here. But any method is okay, so long as the calculations are valid. ∞ X n! b) 10n n=1
In this case, when n gets large (which is the only limit we care about), the factorial in the numerator will start to dominate over the power in the denominator. So we expect this to diverge. As a proof, we can perform a simple ratio test. an = Taking the limit, we obtain
n! 10n
⇒
an 10 = an+1 n+1
an =0 n→∞ an+1 lim
hence the series diverges by the ratio test.
c)
∞ X
1 2n(2n + 1) n=1 We first note that this series behaves like 1/4n2 for large n. As a result, we expect it to converge. To see this, we may consider a simple comparison test 1 1 1 < = an = 2n(2n + 1) 2n · 2n 4 Since the series ζ(2) =
d)
∞ X
P∞
n=1 (1/n
2
1 n2
) converges, this series converges as well.
[n(n + 1)]−1/2
n=1
This series behaves as 1/n for large n. Thus we expect it to diverge. While the square root may be a bit awkward to manipulate, we can actually perform a simple comparison test with the harmonic series 1 1 1 an = p >p = n+1 n(n + 1) (n + 1)(n + 1) Because the harmonic series diverges (and we do not care that the comparison starts with the second term in the harmonic series, and not the first) this series also diverges. e)
∞ X
1 2n + 1 n=0 Since this behaves as 1/2n for large n, the series ought to diverge. We may either compare this with the harmonic series or perform an integral test. Consider the integral test ∞ Z ∞ dx 1 = ln(2x + 1) = ∞ 2x + 1 2 0
0
Thus the series diverges 5.2.8 For what values of p and q will the following series converge?
∞ X
1/ [np (ln n)q ]
n=2
Since the ln n term is not as dominant as the power term np , we may have some idea that the series ought to converge or diverge as the 1/np series. To make this
more precise, we can use Raabe’s test an =
1 p n (ln n)q
an (n + 1)p (ln(n + 1))q = an+1 np (ln n)q p q ln(1 + n1 ) 1 = 1+ 1+ n ln n p q 1 1 = 1+ 1+ + ··· n n ln n p q = 1 + + ··· 1 + + ··· n nln n p q = 1+ + + ··· n n ln n
⇒
Note that we have Taylor (or binomial) expanded the expressions several times. Raabe’s test then yields lim n
n→∞
an q + ··· = p − 1 = lim p + n→∞ an+1 ln n
This gives convergence for p > 1 and divergence for p < 1. For p = 1, Raabe’s test is ambiguous. However, in this case we can perform an integral test. Since 1 p = 1 ⇒ an = n(ln n)q we evaluate
Z 2
∞
dx = x(ln x)q
Z
∞
ln 2
du uq
where we have used the substitution u = ln x. This converges for q > 1 and diverges otherwise. Hence the final result is p > 1, p = 1, p = 1, p < 1,
any q q>1 q≤1 any q
converge converge diverge diverge
5.2.9 Determine the range of convergence for Gauss’s hypergeometric series F (α, β, γ; x) = 1 +
α(α + 1)β(β + 1) 2 αβ x+ x + ··· 1!γ 2!γ(γ + 1)
We first consider non-negative values of x (so that this is a positive series). More or less, this is a power series in x. So as long as α, β, γ are well behaved, this
series ought to converge for x < 1 (just like an ordinary geometric series). To see this (and to prepare for Gauss’ test), we compute the ratio an =
α(α + 1) · · · (α + n − 1)β(β + 1) · · · (β + n − 1) n x n!γ(γ + 1) · · · (γ + n − 1) an (n + 1)(γ + n) −1 ⇒ = x an+1 (α + n)(β + n)
This allows us to begin with the ratio test an (n + 1)(γ + n) −1 = lim x = x−1 n→∞ an+1 n→∞ (α + n)(β + n) lim
Hence the series converges for x < 1 and diverges for x > 1. However, the ratio test is indeterminate for x = 1. This is where we must appeal to Gauss’ test. Setting x = 1, we have an (n + 1)(γ + n) = an+1 (α + n)(β + n) Since this approaches 1 as n → ∞, we may highlight this leading behavior by adding and subtracting 1 an =1+ an+1
(n + 1)(γ + n) −1 (α + n)(β + n)
=1+
(γ − α − β + 1)n + γ − αβ (α + n)(β + n)
We can now see that the fraction approaches (γ −α−β +1)/n as n gets large. This is the h/n behavior that we need to extract for Gauss’ test: an /an+1 = 1 + h/n + B(n)/n2 . In principle, we may add and subtract h/n where h = γ − α − β + 1 in order to obtain an explicit expression for the remainder term B(n)/n2 . However, it should be clear based on a power series expansion that this remainder will indeed behave as ∼ 1/n2 , which is the requirement for applying Gauss’ test. Thus, with h = γ − α − β + 1, we see that the hypergeometric series F (α, β, γ; 1) converges for γ > α + β (h > 1) and diverges otherwise. To summarize, we have proven that for non-negative x, the hypergeometric series converges for x < 1 (any α, β, γ) and x = 1 if γ > α + β, and diverges otherwise. In fact, for negative values of x, we may consider the series for |x|. In this case, we have absolute convergence for |x| < 1 and |x| = 1 if γ > α + β. Based on the ratio test, it is not hard to see that the series also diverges for |x| > 1 (for negative x, each subsequent term gets larger than the previous one). However, there is also conditional convergence for α + β − 1 < γ ≤ α + β (this is harder to show).
5.2.19 Show that the following series is convergent. ∞ X s=0
(2s − 1)!! (2s)!!(2s + 1)
It is somewhat hard to see what happens when s gets large. However, we can perform Raabe’s test as =
(2s − 1)!! (2s)!!(2s + 1)
⇒
as (2s − 1)!! (2s + 2)!!(2s + 3) = × as+1 (2s)!!(2s + 1) (2s + 1)!! (2s − 1)!!(2s + 2)!!(2s + 3) = (2s + 1)!! (2s)!! (2s + 1) (2s + 2)(2s + 3) = (2s + 1)(2s + 1)
By adding and subtracting 1, we obtain as =1+ as+1 Then
lim s
s→∞
(2s + 2)(2s + 3) 6s + 5 −1 =1+ 2 (2s + 1) (2s + 1)2
as −1 as+1
= lim s s→∞
6s + 5 (2s + 1)2
=
3 2
Since this is greater than 1, the series converges. 5.3.1
a) From the electrostatic two hemisphere problem we obtain the series ∞ X s=0
(−1)s (4s + 3)
(2s − 1)!! (2s + 2)!!
Test it for convergence. Since this is an alternating series, we may check if it is monotonic decreasing. Taking the ratio, we see that |as | (4s + 3)(2s − 1)!!(2s + 4)!! (4s + 3)(2s + 4) = = |as+1 | (4s + 7)(2s + 1)!!(2s + 2)!! (4s + 7)(2s + 1) 2 8s + 22s + 12 4s + 5 = = 1 + >1 8s2 + 18s + 7 8s2 + 18s + 7 As a result |as | > |as+1 |
and hence the series converges based on the Leibniz criterion. (Actually, to be careful, we must also show that lims→∞ as = 0. However, I have ignored this subtlety.) b) The corresponding series for the surface charge density is ∞ X s=0
(−1)s (4s + 3)
(2s − 1)!! (2s)!!
Test it for convergence. This series is rather similar to that of part a). However the denominator is ‘missing’ a factor of (2s + 2). This makes the series larger (term by term) than the above. To see whether the terms get too large, we may take the ratio (4s + 3)(2s − 1)!!(2s + 2)!! (4s + 3)(2s + 2) |as | = = |as+1 | (4s + 7)(2s + 1)!! (2s)!! (4s + 7)(2s + 1) 2 8s + 14s + 6 4s + 1 = 2 =1− 2 <1 8s + 18s + 7 8s + 18s + 7 In this case |as | < |as+1 | and the series diverges since the terms get larger as s → ∞.
