TITLE: Convergence to half Euler's constant
QUESTION [6 upvotes]: Euler's constant is defined by $$\gamma=\lim_{N\rightarrow\infty}\sum_{n=1}^N\dfrac1n-\log N.$$
So we can write it as $$1+\dfrac12+\dfrac13+\ldots+\dfrac1n-\log n\rightarrow \gamma.$$
How can we show that $$1+\dfrac13+\dfrac15+\ldots+\dfrac{1}{2n-1}-\dfrac12\log n\rightarrow\dfrac{\gamma}{2}+\log 2?$$
The fractions are similar, but when they skip the even terms, it's not clear how to relate.

REPLY [2 votes]: Adding and substracting the inverses of even integers, one sees that the sequence whose limit you are asking is
$$
\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1{2k}-\frac12\log n=(H_{2n}-\log(2n))-\frac12(H_n-\log n)+\log2,
$$
where $H_k$ is the $k$th harmonic number and where we used the fact that $\log(2n)=\log n+\log2$. Since $H_k-\log k\to\gamma$ when $k\to\infty$, this converges to
$$
\gamma-\frac12\gamma+\log2=\frac12\gamma+\log2.
$$