Putnam 2017, Problem B-4

Evaluate the sum  
\[
\sum_{k=0}^\infty \left( 3 \cdot \frac{\ln(4k+2)}{4k+2} - \frac{\ln(4k+3)}{4k+3} - \frac{\ln(4k+4)}{4k+4} - \frac{\ln(4k+5)}{4k+5} \right)
\]  
\[
= 3 \cdot \frac{\ln 2}{2} - \frac{\ln 3}{3} - \frac{\ln 4}{4} - \frac{\ln 5}{5} + 3 \cdot \frac{\ln 6}{6} - \frac{\ln 7}{7} - \frac{\ln 8}{8} - \frac{\ln 9}{9} + 3 \cdot \frac{\ln 10}{10} - \cdots.
\]  
(As usual, \( \ln x \) denotes the natural logarithm of \( x \).)

Solution:
We prove that the sum equals \( (\log 2)^2 \); as usual, we write \( \log x \) for the natural logarithm of \( x \) instead of \( \ln x \). Note that of the two given expressions of the original sum, the first is absolutely convergent (the summands decay as \( \log(x)/x^2 \)) but the second one is not; we must thus be slightly careful when rearranging terms.  

### First solution
Define \( a_k = \frac{\log k}{k} - \frac{\log(k+1)}{k+1} \). The infinite sum \( \sum_{k=1}^\infty a_k \) converges to \( 0 \) since \( \sum_{k=1}^n a_k \) telescopes to \( -\frac{\log(n+1)}{n+1} \) and this converges to \( 0 \) as \( n \to \infty \). Note that \( a_k > 0 \) for \( k \geq 3 \) since \( \frac{\log x}{x} \) is a decreasing function of \( x \) for \( x > e \), and so the convergence of \( \sum_{k=1}^\infty a_k \) is absolute.  

Write \( S \) for the desired sum. Then since \( 3a_{4k+2} + 2a_{4k+3} + a_{4k+4} = (a_{4k+2} + a_{4k+4}) + 2(a_{4k+2} + a_{4k+3}) \), we have  
\[
S = \sum_{k=0}^\infty (3a_{4k+2} + 2a_{4k+3} + a_{4k+4}) = \sum_{k=1}^\infty a_{2k} + \sum_{k=0}^\infty 2(a_{4k+2} + a_{4k+3}),
\]  
where we are allowed to rearrange the terms in the infinite sum since \( \sum a_k \) converges absolutely. Now \( 2(a_{4k+2} + a_{4k+3}) = \frac{\log(4k+2)}{2k+1} - \frac{\log(4k+4)}{2k+2} = a_{2k+1} + (\log 2)(\frac{1}{2k+1} - \frac{1}{2k+2}) \), and summing over \( k \) gives  
\[
\sum_{k=0}^\infty 2(a_{4k+2} + a_{4k+3}) = \sum_{k=0}^\infty a_{2k+1} + (\log 2) \sum_{k=0}^\infty \frac{(-1)^{k+1}}{k} = \sum_{k=0}^\infty a_{2k+1} + (\log 2)^2.
\]  

Thus  
\[
S = \sum_{k=0}^\infty a_{2k} + \sum_{k=0}^\infty a_{2k+1} + (\log 2)^2 = \sum_{k=0}^\infty a_k + (\log 2)^2 = (\log 2)^2.
\]  

### Second solution
We start with the following observation: for any positive integer \( n \),  
\[
\left. \frac{d}{ds} n^{-s} \right|_{s=1} = -(\log n) n^{-s}.
\]  

(Throughout, we view \( s \) as a real parameter, but see the remark below.) For \( s > 0 \), consider the absolutely convergent series  
\[
L(s) = \sum_{k=0}^\infty \left( 3(4k+2)^{-s} - (4k+3)^{-s} - (4k+4)^{-s} - (4k+5)^{-s} \right);
\]  
in the same range we have  
\[
L'(s) = \sum_{k=0}^\infty \left( 3 \frac{\log(4k+2)}{(4k+2)^s} - \frac{\log(4k+3)}{(4k+3)^s} - \frac{\log(4k+4)}{(4k+4)^s} - \frac{\log(4k+5)}{(4k+5)^s} \right),
\]  
so we may interchange the summation with taking the limit at \( s = 1 \) to equate the original sum with \( -L'(1) \).  

To make further progress, we introduce the Riemann zeta function \( \zeta(s) = \sum_{n=1}^\infty n^{-s} \), which converges absolutely for \( s > 1 \). In that region, we may freely rearrange sums to write  
\[
L(s) + \zeta(s) = 1 + 4(2^{-s} + 6^{-s} + 10^{-s} + \cdots) = 1 + 2^{2-s}(1 + 3^{-s} + 5^{-s} + \cdots) = 1 + 2^{2-s}(\zeta(s) - 2^{-s} - 4^{-s} - \cdots)
\]  
\[
= 1 + 2^{2-s} \zeta(s) - 2^{2-2s} \zeta(s).
\]  

In other words, for \( s > 1 \), we have  
\[
L(s) = 1 + \zeta(s)(-1 + 2^{2-s} - 2^{2-2s}).
\]  

Now recall that \( \zeta(s) - \frac{s}{s-1} \) extends to a \( C^\infty \) function for \( s > 0 \), e.g., by applying Abel summation to obtain  
\[
\zeta(s) - \frac{s}{s-1} = \sum_{n=1}^\infty n(n^{-s} - (n+1)^{-s}) - \frac{s}{s-1} = s \sum_{n=1}^\infty n \int_n^{n+1} x^{-s-1} dx - \frac{s}{s-1} = -s \int_1^\infty (x - \lfloor x \rfloor) x^{-s-1} dx.
\]  

Also by writing \( 2^{2-s} = 2 \exp((1 - s) \log 2) \) and \( 2^{2-2s} = \exp(2(1 - s) \log 2) \), we may use the exponential series to compute the Taylor expansion of  
\[
f(s) = \frac{-1 + 2^{2-s} - 2^{2-2s}}{s - 1}
\]  
at \( s = 1 \); we get  
\[
f(s) = -(\log 2)^2 (s - 1)^2 + O((s - 1)^3).
\]  

Consequently, if we rewrite the previous expression for \( L(s) \) as  
\[
L(s) = 1 + (s - 1) \zeta(s) \cdot \frac{-1 + 2^{2-s} - 2^{2-2s}}{s - 1},
\]  
then we have an equality of \( C^\infty \) functions for \( s > 1 \), and hence (by continuity) an equality of Taylor series about \( s = 1 \). That is,  
\[
L(s) = 1 - (\log 2)^2 (s - 1) + O((s - 1)^2),
\]  
which yields the desired result.  

### Remark
The use of series \( \sum_{n=1}^\infty c_n n^{-s} \) as functions of a real parameter \( s \) dates back to Euler, who observed that the divergence of \( \zeta(s) \) as \( s \to 1 \) gives a proof of the infinitude of primes distinct from Euclid’s approach, and Dirichlet, who upgraded this idea to prove his theorem on the distribution of primes across arithmetic progressions. It was Riemann who introduced the idea of viewing these series as functions of a complex parameter, thus making it possible to use the tools of complex analysis (e.g., the residue theorem) and leading to the original proof of the prime number theorem by Hadamard and de la Vallée Poussin.  

In the language of complex analysis, one may handle the convergence issues in the second solution in a different way: use the preceding calculation to establish the equality  
\[
L(s) = 1 + \zeta(s)(-1 + 2^{2-s} - 2^{2-2s})
\]  
for \( \operatorname{Re}(s) > 1 \), then observe that both sides are holomorphic for \( \operatorname{Re}(s) > 0 \) and so the equality extends to that larger domain.