Section 7

Lambert Series and Related Topics

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Problem 8.65

From the identity

ζ(s)∑n=1∞ann−s=∑n=1∞Ann−s \zeta (s)\sum _{n=1}^\infty a_nn^{-s} = \sum _{n=1}^\infty A_nn^{-s}

it follows that

∑n=1∞anxn1−xn=∑n=1∞Anxn \sum _{n=1}^\infty \frac{a_nx^n}{1-x^n} = \sum _{n=1}^\infty A_nx^n

and conversely, where a1,a2,a3,…,A1,A2,A3,…a_1,a_2,a_3,\ldots ,A_1,A_2,A_3,\ldots are constants. (The series occurring on the left-hand side of the second equation is called a Lambert series.)

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Problem 8.66

The two identities

ζ(s)(1−21−s)∑n=1∞ann−s=∑n=1∞Bnn−s,∑n=1∞anxn1+xn=∑n=1∞Bnxn \zeta (s)(1-2^{1-s})\sum _{n=1}^\infty a_nn^{-s} = \sum _{n=1}^\infty B_nn^{-s} , \qquad \sum _{n=1}^\infty \frac{a_nx^n}{1+x^n} = \sum _{n=1}^\infty B_nx^n

are equivalent.

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Problem 8.67

Assume that between the numbers a1,a2,a3,…a_1,a_2,a_3,\ldots, and A1,A2,A3,…A_1,A_2,A_3,\ldots the same relation subsists as in 65. Then

∏n=1∞(nxsin⁡xn)an=∏n=1∞(1−x2n2π2)An. \prod _{n=1}^\infty \left(\frac nx \sin \frac xn\right)^{a_n} = \prod _{n=1}^\infty \left(1-\frac{x^2}{n^2\pi ^2}\right)^{A_n} .
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Problem 8.68

Assume that between the numbers a1,a2,a3,…a_1,a_2,a_3,\ldots and B1,B2,B3,…B_1,B_2,B_3,\ldots the same relation subsists as in 66. Then

∏n=1∞(x2ncot⁡x2n)an=∏n=1∞(1−x2n2π2)Bn. \prod _{n=1}^\infty \left(\frac{x}{2n}\cot \frac{x}{2n}\right)^{a_n} = \prod _{n=1}^\infty \left(1-\frac{x^2}{n^2\pi ^2}\right)^{B_n} .
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Problem 8.68.1

Define

f(x)+f(2x)+⋯+f(nx)+⋯=F(x). f(x)+f(2x)+\cdots +f(nx)+\cdots = F(x) .

Then, with the notation of 65,

∑n=1∞anF(nx)=∑n=1∞Anf(nx). \sum _{n=1}^\infty a_nF(nx) = \sum _{n=1}^\infty A_nf(nx) .

(How is this result related to 65 and 67?)

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Problem 8.68.2

Define

f(x)−f(2x)+⋯+(−1)n−1f(nx)+⋯=G(x). f(x)-f(2x)+\cdots +(-1)^{n-1}f(nx)+\cdots = G(x) .

Then, with the notation of 66,

∑n=1∞anG(nx)=∑n=1∞Bnf(nx). \sum _{n=1}^\infty a_nG(nx) = \sum _{n=1}^\infty B_nf(nx) .

(How is this result related to 66 and 68?)

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Problem 8.68.3

Each of the following two equations implies the other:

F(x)=∑n=1∞f(nx),f(x)=∑n=1∞μ(n)F(nx). F(x) = \sum _{n=1}^\infty f(nx) , \qquad f(x) = \sum _{n=1}^\infty \mu (n)F(nx) .
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Problem 8.69

Show that

∑n=1∞μ(n)xn1−xnand∑n=1∞φ(n)xn1−xn \sum _{n=1}^\infty \frac{\mu (n)x^n}{1-x^n} \qquad \text{and} \qquad \sum _{n=1}^\infty \frac{\varphi (n)x^n}{1-x^n}

are rational functions of xx. What functions?

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Problem 8.70
∑n=1∞λ(n)xn1−xn=x+x4+x9+x16+x25+⋯ . \sum _{n=1}^\infty \frac{\lambda (n)x^n}{1-x^n} = x+x^4+x^9+x^{16}+x^{25}+\cdots .
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Problem 8.71

Show that

∑n=1∞μ(n)xn1+xn=x−2x2,∑n=1∞φ(n)xn1+xn=x1+x2(1−x2)2. \sum _{n=1}^\infty \frac{\mu (n)x^n}{1+x^n} = x-2x^2 , \qquad \sum _{n=1}^\infty \frac{\varphi (n)x^n}{1+x^n} = x\frac{1+x^2}{(1-x^2)^2} .
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Problem 8.72
∑n=1∞λ(n)xn1+xn=x−2x2+x4−2x8+x9+x16−2x18+x25−⋯=∑n=1∞bnxn, \sum _{n=1}^\infty \lambda (n)\frac{x^n}{1+x^n} = x-2x^2+x^4-2x^8+x^9+x^{16}-2x^{18}+x^{25}-\cdots = \sum _{n=1}^\infty b_nx^n ,

where bn=1b_n=1 if nn is a square, bn=−2b_n=-2 if nn is twice a square, and in all other cases bn=0b_n=0.

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Problem 8.72.1
∏1∞(1−xn)μ(n)n=e−x,∏1∞(1−xn)φ(n)n=e−x1−x. \prod _{1}^\infty \left(1-x^n\right)^{\frac{\mu (n)}{n}} = e^{-x} , \qquad \prod _{1}^\infty \left(1-x^n\right)^{\frac{\varphi (n)}{n}} = e^{-\frac{x}{1-x}} .
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Problem 8.73
∑n=1∞Φ(n)xn1−xn=x1+x(1−x)3,∑n=1∞Φ(n)xn1+xn=x1+2x+6x2+2x3+x4(1−x2)3. \sum _{n=1}^\infty \Phi (n)\frac{x^n}{1-x^n} = x\frac{1+x}{(1-x)^3} , \qquad \sum _{n=1}^\infty \Phi (n)\frac{x^n}{1+x^n} = x\frac{1+2x+6x^2+2x^3+x^4}{(1-x^2)^3} .
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Remark.
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Cf. Problem 64.

Problem 8.74
∑n=1∞τ(n)xn=∑n=1∞xn1−xn=x1+x1−x+x41+x21−x2+x91+x31−x3+⋯ . \sum _{n=1}^\infty \tau (n)x^n = \sum _{n=1}^\infty \frac{x^n}{1-x^n} = x\frac{1+x}{1-x}+x^4\frac{1+x^2}{1-x^2}+x^9\frac{1+x^3}{1-x^3}+\cdots .
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Problem 8.74.1
∑n=1∞δ(n)xn=x1−x−x31−x3+x51−x5−x71−x7+⋯ . \sum _{n=1}^\infty \delta (n)x^n = \frac{x}{1-x}-\frac{x^3}{1-x^3}+\frac{x^5}{1-x^5}-\frac{x^7}{1-x^7}+\cdots .
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Remark.
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Cf. Problem 58.7.

Problem 8.75
∑n=1∞σ(n)xn=∑n=1∞nxn1−xn=x(1−x)2+x2(1−x2)2+x3(1−x3)2+⋯ . \sum _{n=1}^\infty \sigma (n)x^n = \sum _{n=1}^\infty n\frac{x^n}{1-x^n} = \frac{x}{(1-x)^2} + \frac{x^2}{(1-x^2)^2} + \frac{x^3}{(1-x^3)^2} + \cdots .
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Problem 8.75.1
∑n=1∞σ(n)xn=x+2x2−5x5−7x7++−−⋯1−x−x2+x5+x7−−++⋯. \sum _{n=1}^\infty \sigma (n)x^n = \frac{x+2x^2-5x^5-7x^7+{+}-{-}\cdots }{1-x-x^2+x^5+x^7-{-}+{+}\cdots } .
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Remark.
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Cf. Problem 54 of Part I.

Problem 8.75.2
3∑n=1∞σ(n)xn=3x−15x3+42x6−90x10+⋯1−3x+5x3−7x6+9x10−⋯. 3\sum _{n=1}^\infty \sigma (n)x^n = \frac{3x-15x^3+42x^6-90x^{10}+\cdots }{1-3x+5x^3-7x^6+9x^{10}-\cdots } .
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Problem 8.75.3
σ(n)=σ(n−1)+σ(n−2)−σ(n−5)−σ(n−7)++−−⋯ . \sigma (n) = \sigma (n-1)+\sigma (n-2)-\sigma (n-5)-\sigma (n-7)+{+}-{-}\cdots .

The terms on the right-hand side are of the form

(−1)k−1σ(n−3k2±k2),where1≤3k2±k2≤n. (-1)^{k-1}\sigma \left(n-\frac{3k^2\pm k}{2}\right) , \qquad \text{where} \quad 1\leq \frac{3k^2\pm k}{2}\leq n .

Where the symbol σ(0)\sigma (0), till now undefined, occurs, substitute for it nn.

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Problem 8.75.4
σ(n)=3σ(n−1)−5σ(n−3)+7σ(n−6)−9σ(n−10)+−⋯ . \sigma (n) = 3\sigma (n-1)-5\sigma (n-3)+7\sigma (n-6)-9\sigma (n-10)+{-}\cdots .

The terms on the right-hand side are of the form

(−1)k−1(2k+1)σ(n−k(k+1)2),where1≤k(k+1)2≤n. (-1)^{k-1}(2k+1)\sigma \left(n-\frac{k(k+1)}{2}\right) , \qquad \text{where} \quad 1\leq \frac{k(k+1)}{2}\leq n .

Where the symbol σ(0)\sigma (0) occurs, substitute for it n/3n/3.

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Problem 8.75.5

Using the definition of p(n)p(n), the number of partitions of nn, in Problem 20.1 of Part I (Appendix), prove that

σ(n)+σ(n−1)+2σ(n−2)+⋯+p(n−1)σ(1)=∑k=0n−1p(k)σ(n−k)=np(n). \sigma (n)+\sigma (n-1)+2\sigma (n-2)+\cdots +p(n-1)\sigma (1) = \sum _{k=0}^{n-1} p(k)\sigma (n-k) = np(n) .
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Problem 8.76

The expression

11−q+p1−qx+p21−qx2+p31−qx3+⋯ \frac{1}{1-q}+\frac{p}{1-qx}+\frac{p^2}{1-qx^2}+\frac{p^3}{1-qx^3}+\cdots

is unaltered, if pp and qq are interchanged.

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Problem 8.77
x1+x2+x21+x4+x31+x6+x41+x8+⋯=x1−x−x31−x3+x51−x5−x71−x7+⋯ . \frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x^6}+\frac{x^4}{1+x^8}+\cdots = \frac{x}{1-x}-\frac{x^3}{1-x^3}+\frac{x^5}{1-x^5}-\frac{x^7}{1-x^7}+\cdots .
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Problem 8.78
x1−x=x1−x2+x21−x4+x41−x8+x81−x16+⋯ \frac{x}{1-x} = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\frac{x^8}{1-x^{16}}+\cdots =x1+x+2x21+x2+4x41+x4+8x81+x8+⋯ . = \frac{x}{1+x}+\frac{2x^2}{1+x^2}+\frac{4x^4}{1+x^4}+\frac{8x^8}{1+x^8}+\cdots .
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