- Show that
\[ 1 + \frac {1}{3} + \frac {1}{5} + \dots + \frac {1}{2 n - 1} - \frac {1}{2} \log n \rightarrow \frac {\gamma}{2} + \log 2, \]
where \(\gamma\) is Euler’s constant.
Solution
Let \[ H_n=1+\frac12+\cdots+\frac1n. \] The odd terms up to \(2n-1\) satisfy \[ 1+\frac13+\cdots+\frac1{2n-1} =H_{2n}-\frac12H_n. \] Hence \[ 1+\frac13+\cdots+\frac1{2n-1}-\frac12\log n =H_{2n}-\frac12H_n-\frac12\log n. \] Using \[ H_m=\log m+\gamma+o(1), \] we obtain \[ \log(2n)+\gamma-\frac12(\log n+\gamma)-\frac12\log n+o(1) =\log2+\frac\gamma2+o(1). \] Therefore \[ \boxed{1+\frac13+\cdots+\frac1{2n-1}-\frac12\log n \longrightarrow \frac\gamma2+\log2}. \]