- This problem provides further estimates for \(\zeta\) and \(\zeta'\) near \(\operatorname{Re}(s) = 1\).
Use Proposition 2.5 and its corollary to prove \[ \zeta(s) = \sum_{1 \leq n < N} n^{-s} + \frac{N^{1-s}}{s-1} + \sum_{n \geq N} \delta_n(s) \] for every integer \(N \geq 2\), whenever \(\operatorname{Re}(s) > 0\), where \(\delta_n(s) = n^{-s} - \int_n^{n+1} x^{-s}\,dx\).
Show that \(|\zeta(1+it)| = O(\log |t|)\) as \(|t| \to \infty\) by using the previous result with \(N = \lfloor |t| \rfloor\).
Show that \(|\zeta'(1+it)| = O((\log |t|)^2)\) as \(|t| \to \infty\).
Show that if \(t \neq 0\) and \(t\) is fixed, then the partial sums of the series \(\sum_{n=1}^\infty n^{-1-it}\) are bounded, but the series does not converge.
Solution
1. Part 1(a): Expansion for \(\zeta(s)\) with truncation parameter \(N\).
1.1. For \(\Re(s) > 1\), the Dirichlet series converges absolutely:\[\zeta(s) = \sum_{n=1}^\infty n^{-s} = \sum_{n=1}^{N-1} n^{-s} + \sum_{n=N}^\infty n^{-s}.\]
1.2. For each \(n \ge N\), write \(n^{-s} = \int_n^{n+1} x^{-s}\,dx + \delta_n(s)\), where \(\delta_n(s) = \int_n^{n+1} (n^{-s} - x^{-s})\,dx\).
1.3. Summing the integral terms gives\[\sum_{n=N}^\infty \int_n^{n+1} x^{-s}\,dx = \int_N^\infty x^{-s}\,dx = \left[ \frac{x^{1-s}}{1-s} \right]_N^\infty = \frac{N^{1-s}}{s-1},\]which converges for \(\Re(s) > 1\).
1.4. Thus for \(\Re(s) > 1\),\[\zeta(s) = \sum_{1 \le n < N} n^{-s} + \frac{N^{1-s}}{s-1} + \sum_{n \ge N} \delta_n(s).\]
1.5. By the Mean Value Theorem applied to \(f(x) = x^{-s}\) on \([n, n+1]\), \(|n^{-s} - x^{-s}| \le |s| n^{-\Re(s)-1}\), so\[|\delta_n(s)| \le |s| n^{-\Re(s)-1}.\]The series \(\sum_{n \ge N} \delta_n(s)\) converges locally uniformly for \(\Re(s) > 0\), defining a holomorphic function on \(\Re(s) > 0\).
1.6. By analytic continuation, the identity holds for all \(\Re(s) > 0\) with \(s \neq 1\).
2. Part 1(b): Bound \(|\zeta(1+it)| = O(\log |t|)\) as \(|t| \to \infty\).
2.1. Set \(s = 1+it\) with \(|t| \ge 2\), and choose \(N = \lfloor |t| \rfloor \ge 2\).
2.2. In the expansion from 1:\[\left| \sum_{1 \le n < N} n^{-1-it} \right| \le \sum_{n=1}^{N-1} \frac{1}{n} \le 1 + \int_1^N \frac{dx}{x} = 1 + \log N \le 1 + \log |t|.\]
2.3. For the pole term:\[\left| \frac{N^{1-s}}{s-1} \right| = \left| \frac{N^{-it}}{it} \right| = \frac{1}{|t|} \le \frac{1}{2} = O(1).\]
2.4. For the tail: using \(|\delta_n(1+it)| \le |1+it| n^{-2} \le (1+|t|) n^{-2}\),\[\left| \sum_{n=N}^\infty \delta_n(1+it) \right| \le (1+|t|) \sum_{n=N}^\infty \frac{1}{n^2} \le (1+|t|) \frac{1}{N-1} \le \frac{1+|t|}{|t|-1} \le 3 = O(1).\]
2.5. Summing the three bounds gives \(|\zeta(1+it)| \le \log |t| + O(1) = O(\log |t|)\).
3. Part 1(c): Bound \(|\zeta'(1+it)| = O((\log |t|)^2)\) as \(|t| \to \infty\).
3.1. Differentiating the identity from 1 with respect to \(s\):\[\zeta'(s) = -\sum_{1 \le n < N} n^{-s} \log n - \frac{N^{1-s} \log N}{s-1} - \frac{N^{1-s}}{(s-1)^2} + \sum_{n=N}^\infty \delta_n'(s).\]
3.2. At \(s = 1+it\) with \(N = \lfloor |t| \rfloor\):\[\left| \sum_{1 \le n < N} n^{-1-it} \log n \right| \le \sum_{n=1}^{N-1} \frac{\log n}{n} \le \int_1^N \frac{\log x}{x}\,dx + O(1) = \frac{1}{2} (\log N)^2 + O(1) \le \frac{1}{2}(\log |t|)^2 + O(1).\]
3.3. The terms involving \(N^{1-s}\) satisfy\[\left| \frac{N^{-it} \log N}{it} \right| = \frac{\log N}{|t|} = O(1), \qquad \left| \frac{N^{-it}}{(it)^2} \right| = \frac{1}{|t|^2} = O(1).\]
3.4. Differentiating \(\delta_n(s) = \int_n^{n+1} (n^{-s} - x^{-s})\,dx\) gives \(|\delta_n'(1+it)| \le C |t| \frac{\log n}{n^2}\), so\[\left| \sum_{n=N}^\infty \delta_n'(1+it) \right| \le C |t| \sum_{n=N}^\infty \frac{\log n}{n^2} = O\left(|t| \frac{\log N}{N}\right) = O(\log |t|).\]
3.5. Therefore \(|\zeta'(1+it)| = O((\log |t|)^2)\).
4. Part 1(d): Partial sums are bounded but do not converge for \(t \neq 0\).
4.1. Let \(S_m(t) = \sum_{n=1}^m n^{-1-it}\). Applying 1 with \(s = 1+it\) and \(N = m+1\):\[S_m(t) = \zeta(1+it) - \frac{(m+1)^{-it}}{it} - \sum_{n=m+1}^\infty \delta_n(1+it).\]
4.2. The tail satisfies \(\left| \sum_{n=m+1}^\infty \delta_n(1+it) \right| \le (1+|t|) \frac{1}{m} \le 1+|t|\) for all \(m \ge 1\).
4.3. Since \(|\zeta(1+it)|\) is a constant for fixed \(t\), and \(|(m+1)^{-it}/(it)| = 1/|t|\) is constant, \(|S_m(t)| \le |\zeta(1+it)| + \frac{1}{|t|} + (1+|t|)\) is uniformly bounded in \(m\).
4.4. If \(S_m(t)\) converged to \(\zeta(1+it)\) as \(m \to \infty\), then \((m+1)^{-it} = e^{-it \log(m+1)}\) would converge as \(m \to \infty\). But for \(t \neq 0\), the sequence \(e^{-it \log(m+1)}\) oscillates around the unit circle and has no limit.
4.5. Hence the series \(\sum_{n=1}^\infty n^{-1-it}\) diverges.
5. Q.E.D.
Steps 1 to 4 prove parts (a) to (d).