Other closed forms of $\lim_{k\rightarrow\infty}\left(\sum_{i=0}^k\frac{1}{2i+1}-\sum_{i=1}^k\frac{1}{2i}\right)$

Question

It is known that $$\lim_{k\rightarrow\infty}\left(\sum_{i=0}^k\frac{1}{2i+1}-\sum_{i=1}^k\frac{1}{2i}\right)=\beta$$converges. I wonder if there are any other closed forms for this limit. At first I thought this was equivalent to $$\int_0^\infty\frac{1}{\lfloor2x+3\rfloor}-\frac{1}{\lfloor2x+2\rfloor}dx$$but it wasn't. I thought of this by using similar reasoning to why $$\gamma=\int_1^\infty\left(\frac{1}{\lfloor x\rfloor}-\frac{1}{x}\right)dx$$

Please don't write down anything obvious like $\beta=\beta+a-a$ for some complex number $a$.

Edit: Thanks to @Empy2's comment, $\beta=\ln2$.

Answer 1

$$ \sum_{i=0}^k\frac{1}{2i+1}-\sum_{i=1}^k\frac{1}{2i}=\left(\frac{1}{2}H_{k+1/2}+\ln2\right)-\frac{1}{2}H_{k}=\frac12\left(H_{k+1/2}-H_k\right)+\ln2\to\ln2 \quad \text{as }k\to\infty $$