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Number Theory Level 5

A = ∑ n = 1 ∞ ( − 1 ) n d ( n ) n 2 A=\sum _{n=1}^{\infty }\frac{\left(-1\right)^nd\left(n\right)}{n^2}

Find ⌊ 10000 A ⌋ \left\lfloor 10000A \right\rfloor

Note : d ( n ) d(n) denotes the number of positive integer divisors of n n .

Context


The answer is -3383.

Julian Poon by Julian Poon , 20, Singapore

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1 solution

Jake Lai
Dec 19, 2015

Observe that ∑ n = 1 ∞ f ( n ) = ∑ k = 0 ∞ ∑ n = 0 ∞ f ( 2 k ( 2 n + 1 ) ) \displaystyle \sum_{n=1}^\infty f(n) = \sum_{k=0}^\infty \sum_{n=0}^\infty f(2^k(2n+1)) . This fact is significant since 2 k 2^k and 2 n + 1 2n+1 are always coprime, which makes manipulation simple if f is a multiplicative function.

Since d d is multiplicative,

∑ n = 1 ∞ ( − 1 ) n d ( n ) n 2 = ∑ k = 0 ∞ ∑ n = 0 ∞ ( − 1 ) 2 k ( 2 n + 1 ) d ( 2 k ( 2 n + 1 ) ) [ 2 k ( 2 n + 1 ) ] 2 = ∑ k = 1 ∞ ∑ n = 0 ∞ d ( 2 k ) d ( 2 n + 1 ) 4 k ( 2 n + 1 ) 2 − ∑ n = 0 ∞ d ( 2 n + 1 ) ( 2 n + 1 ) 2 = ∑ k = 1 ∞ k + 1 4 k ∑ n = 0 ∞ d ( 2 n + 1 ) ( 2 n + 1 ) 2 − ∑ n = 0 ∞ d ( 2 n + 1 ) ( 2 n + 1 ) 2 = ∑ n = 0 ∞ d ( 2 n + 1 ) ( 2 n + 1 ) 2 [ − 1 + ∑ k = 1 ∞ k + 1 4 k ] . \begin{aligned} \sum_{n=1}^\infty \frac{(-1)^nd(n)}{n^2} &= \sum_{k=0}^\infty \sum_{n=0}^\infty \frac{(-1)^{2^k(2n+1)}d(2^k(2n+1))}{[2^k(2n+1)]^2} \\ &= \sum_{k=1}^\infty \sum_{n=0}^\infty \frac{d(2^k)d(2n+1)}{4^k(2n+1)^2} - \sum_{n=0}^\infty \frac{d(2n+1)}{(2n+1)^2} \\ &= \sum_{k=1}^\infty \frac{k+1}{4^k} \sum_{n=0}^\infty \frac{d(2n+1)}{(2n+1)^2} - \sum_{n=0}^\infty \frac{d(2n+1)}{(2n+1)^2} \\ & = \sum_{n=0}^\infty \frac{d(2n+1)}{(2n+1)^2} \left[ -1 + \sum_{k=1}^\infty \frac{k+1}{4^k} \right]. \end{aligned}

Let S = ∑ n = 0 ∞ d ( 2 n + 1 ) ( 2 n + 1 ) 2 \displaystyle S = \sum_{n=0}^\infty \frac{d(2n+1)}{(2n+1)^2} and T = ∑ k = 1 ∞ k + 1 4 k \displaystyle T = \sum_{k=1}^\infty \frac{k+1}{4^k} for convenience's sake. Thus, ∑ n = 1 ∞ ( − 1 ) n d ( n ) n 2 = S ( T − 1 ) \displaystyle \sum_{n=1}^\infty \frac{(-1)^nd(n)}{n^2} = S(T-1) .

Now, noting that ζ 2 ( s ) = ∑ n = 1 ∞ d ( n ) n s \displaystyle \zeta^2(s) = \sum_{n=1}^\infty \frac{d(n)}{n^s} , we apply the fact mentioned in the beginning once again:

S = ∑ n = 0 ∞ d ( 2 n + 1 ) ( 2 n + 1 ) 2 = ∑ n = 1 ∞ d ( n ) n 2 − ∑ n = 1 ∞ d ( 2 n ) ( 2 n ) 2 = ζ 2 ( 2 ) − ∑ k = 1 ∞ ∑ n = 0 ∞ d ( 2 k ( 2 n + 1 ) ) [ 2 k ( 2 n + 1 ) ] 2 = ζ 2 ( 2 ) − ∑ k = 1 ∞ k + 1 4 k ∑ n = 0 ∞ d ( 2 n + 1 ) ( 2 n + 1 ) 2 = ζ 2 ( 2 ) − T S . \begin{aligned} S &= \sum_{n=0}^\infty \frac{d(2n+1)}{(2n+1)^2} \\ &= \sum_{n=1}^\infty \frac{d(n)}{n^2} - \sum_{n=1}^\infty \frac{d(2n)}{(2n)^2} \\ &= \zeta^2(2) - \sum_{k=1}^\infty \sum_{n=0}^\infty \frac{d(2^k(2n+1))}{[2^k(2n+1)]^2} \\ &= \zeta^2(2) - \sum_{k=1}^\infty \frac{k+1}{4^k} \sum_{n=0}^\infty \frac{d(2n+1)}{(2n+1)^2} \\ &= \zeta^2(2) - TS. \end{aligned}

Hence, adding TS on both ends of the above equation, we get that S ( T + 1 ) = ζ 2 ( 2 ) = π 4 36 S(T+1) = \zeta^2(2) = \dfrac{\pi^4}{36} .

A quick computation (eg by AGM) reveals that T = 7/9, so combining all our results, we have that

∑ n = 1 ∞ ( − 1 ) n d ( n ) n 2 = S ( T − 1 ) = π 4 36 T − 1 T + 1 = − π 4 288 \sum_{n=1}^\infty \frac{(-1)^nd(n)}{n^2} = S(T-1) = \frac{\pi^4}{36} \frac{T-1}{T+1} = \boxed{-\dfrac{\pi^4}{288}}

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For the sake of completeness, I will present a proof of the identity ζ 2 ( s ) = ∑ n = 1 ∞ d ( n ) n s \displaystyle \zeta^2(s) = \sum_{n=1}^\infty \frac{d(n)}{n^s} .

How do we find a simple way to express the product of two Dirichlet series as a single Dirichlet series? Let D ( h , s ) = ∑ n = 1 ∞ h ( n ) n s \displaystyle \mathfrak{D}(h,s) = \sum_{n=1}^\infty \frac{h(n)}{n^s} for some h : N → C h:\mathbb{N} \to \mathbb{C} and s ∈ C s \in \mathbb{C} . Let us examine the product closely.

D ( f , s ) D ( g , s ) = [ ∑ a = 1 ∞ f ( a ) a s ] [ ∑ b = 1 ∞ f ( b ) b s ] = ∑ a = 1 ∞ ∑ b = 1 ∞ f ( a ) g ( b ) ( a b ) s = ∑ n = 1 ∞ ∑ a b = n f ( a ) g ( b ) n s = ∑ n = 1 ∞ 1 n s ∑ a b = n f ( a ) g ( b ) \begin{aligned} \mathfrak{D}(f,s) \ \mathfrak{D}(g,s) &= \left[ \sum_{a=1}^\infty \frac{f(a)}{a^s} \right] \left[ \sum_{b=1}^\infty \frac{f(b)}{b^s} \right] \\ &= \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{f(a)g(b)}{(ab)^s} \\ &= \sum_{n=1}^\infty \sum_{ab=n} \frac{f(a)g(b)}{n^s} \\ &= \sum_{n=1}^\infty \frac{1}{n^s} \sum_{ab=n} f(a)g(b) \end{aligned}

Now, the last expression appears (and is) in the form of a Dirichlet series, which is well-defined as a function of n! We define ( f ∗ g ) ( n ) = ∑ a b = n f ( a ) g ( b ) = ∑ d ∣ n f ( d ) g ( n / d ) \displaystyle (f*g)(n) = \sum_{ab=n} f(a)g(b) = \sum_{d|n} f(d)g(n/d) . (The operator ∗ * is known as the Dirichlet convolution, and is important in the field of analytic number theory.)

Thus, we have a simple way of expressing the product of two Dirichlet series as a single series:

D ( f , s ) D ( g , s ) = ∑ n = 1 ∞ 1 n s ∑ a b = n f ( a ) g ( b ) = ∑ n = 1 ∞ ( f ∗ g ) ( n ) n s = D ( f ∗ g , s ) . \begin{aligned} \mathfrak{D}(f,s) \ \mathfrak{D}(g,s) &= \sum_{n=1}^\infty \frac{1}{n^s} \sum_{ab=n} f(a)g(b) \\ &= \sum_{n=1}^\infty \frac{(f*g)(n)}{n^s} \\ &= \mathfrak{D}(f*g,s). \end{aligned}

Now, we can prove that D ( d , s ) = ζ 2 ( s ) \displaystyle \mathfrak{D}(d,s) = \zeta^2(s) . First, note that d ( n ) = ∑ d ∣ n 1 = ( 1 ∗ 1 ) ( n ) \displaystyle d(n) = \sum_{d|n} 1 = (\mathbf{1}*\mathbf{1})(n) , where 1 ( n ) = 1 \mathbf{1}(n) = 1 for all n. We then immediately have

D ( d , s ) = D ( 1 ∗ 1 , s ) = D ( 1 , s ) 2 = ζ 2 ( s ) . \mathfrak{D}(d,s) = \mathfrak{D}(\mathbf{1}*\mathbf{1},s) = \mathfrak{D}(\mathbf{1},s)^2 = \zeta^2(s).

Jake Lai - 5 years, 5 months ago

Why aren't you using the Euler product to expand the Dirichlet series? It makes the problem (almost) trivial. The actual computations are essentially the same, but the Euler product makes it much easier to write down.

Otto Bretscher - 5 years, 5 months ago

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Can you explain in greater detail? Both Jake and I don't quite understand what you're talking about.

Pi Han Goh - 5 years, 5 months ago

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Let's assume, as Jake does in his original post, that we know that ∑ d ( n ) n 2 = ζ 2 ( 2 ) \sum\frac{d(n)}{n^2}=\zeta^2(2) . Using an Euler sum, we can write ∑ d ( n ) n 2 = P × ( 1 + 2 4 + 3 4 2 + . . . + k + 1 4 k + . . . ) \sum\frac{d(n)}{n^2}=P\times (1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{k+1}{4^k}+...) = 16 9 P =\frac{16}{9}P since ∑ k = 1 ∞ k + 1 4 k = 7 9 \sum_{k=1}^{\infty}\frac{k+1}{4^k}=\frac{7}{9} . Here, P P incorporates the odd primes.

Now ∑ ( − 1 ) n + 1 d ( n ) n 2 = P × ( 1 − 2 4 − 3 4 2 − . . . − k + 1 4 k − . . . ) = 2 9 P = 1 8 ζ 2 ( 2 ) \sum\frac{(-1)^{n+1}d(n)}{n^2}=P\times (1-\frac{2}{4}-\frac{3}{4^2}-...-\frac{k+1}{4^k}-...)=\frac{2}{9}P=\frac{1}{8}\zeta^2(2) = π 4 288 \frac{\pi^4}{288} .

As I said, it is essentially the same thing, but written much more succinctly.

Otto Bretscher - 5 years, 5 months ago

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