Show that there exists a constant \(\alpha\in[0,1]\) such that \[ \sum_{k=1}^n\frac1{k+\alpha} =\log n+o(1) \qquad(n\to\infty). \]
Solution
There is an \(\alpha\in(0,1)\) for which \[ \sum_{k=1}^n\frac1{k+\alpha}-\log n\longrightarrow0. \] We obtain it by varying continuously between the ordinary harmonic sum and its unit shift.
1. For \(\alpha\in[0,1]\),\[ \sum_{k=1}^n\frac1{k+\alpha}-\log n =\left(H_n-\log n\right) -\sum_{k=1}^n\frac{\alpha}{k(k+\alpha)}, \]where \(H_n=\sum_{k=1}^n1/k\).
For each \(k\), \[ \frac1k-\frac1{k+\alpha} =\frac{\alpha}{k(k+\alpha)}. \] Summing this identity from \(1\) to \(n\) and rearranging gives the displayed formula.
2. The limit\[ L(\alpha) =\lim_{n\to\infty} \left(\sum_{k=1}^n\frac1{k+\alpha}-\log n\right) \]exists for every \(\alpha\in[0,1]\), and\[ L(\alpha) =\gamma- \sum_{k=1}^{\infty}\frac{\alpha}{k(k+\alpha)}. \]
Euler’s constant is \[ \gamma =\lim_{n\to\infty}(H_n-\log(n+1)). \] Since \[ \log(n+1)-\log n\longrightarrow0, \] we also have \[ H_n-\log n\longrightarrow\gamma. \]
For \(0\le\alpha\le1\), \[ 0\le\frac{\alpha}{k(k+\alpha)}\le\frac1{k^2}. \] Thus the series \[ \sum_{k=1}^{\infty}\frac{\alpha}{k(k+\alpha)} \] converges uniformly on \([0,1]\) by the Weierstrass \(M\)-test. Taking limits in 1 gives the formula for \(L\).
3. The function \(L:[0,1]\to\mathbb R\) is continuous.
2. Its sum is therefore continuous, and so is\(L\).
4. Euler’s constant satisfies\[ 0<\gamma<1. \]
For every \(k\ge1\), \[ \int_k^{k+1}\frac{dx}{x}<\frac1k, \] so \[ H_n-\log(n+1)>0, \] and hence \(\gamma\ge0\). The inequality is strict in the limit, for example because already the first summand \[ 1-\log2 \] is positive.
For the upper bound, for \(k\ge2\), \[ \frac1k<\int_{k-1}^{k}\frac{dx}{x}. \] Hence \[ \sum_{k=2}^n\frac1k<\log n, \] and therefore \[ H_n-\log(n+1) <1+\log n-\log(n+1)<1. \] Passing to the limit gives \(\gamma<1\).
5. One has\[ L(0)>0 \qquad\text{and}\qquad L(1)<0. \]
At \(\alpha=0\), 2 gives \[ L(0)=\gamma>0. \] At \(\alpha=1\), \[ \sum_{k=1}^{\infty}\frac1{k(k+1)} =\sum_{k=1}^{\infty}\left(\frac1k-\frac1{k+1}\right) =1. \] Thus \[ L(1)=\gamma-1<0 \] by 4.
6. There exists \(\alpha\in(0,1)\) such that \(L(\alpha)=0\).
By 3, \(L\) is continuous on \([0,1]\), and by 5 it changes sign between the endpoints. The intermediate value theorem therefore gives some \[ \alpha\in(0,1) \] with \(L(\alpha)=0\).
7. This \(\alpha\) has the required asymptotic property.
By the definition of \(L\) and 6, \[ \sum_{k=1}^n\frac1{k+\alpha}-\log n \longrightarrow0. \] Equivalently, \[ \sum_{k=1}^n\frac1{k+\alpha} =\log n+o(1). \]