Prove that $\lim_{x\rightarrow\infty}\sum_{k=0}^x\frac{1}{2k+1}-\sum_{k=1}^x\frac{1}{2k}$ converges

Question

I thought up of the limit in the title by playing around with the limit definition of Euler's constant. I think I was able to prove that it is convergent. Is the following proof correct?

Let $H_e(x)=\sum_{k=1}^x\frac{1}{2k}$ and $H_o(x)=\sum_{k=0}^x\frac{1}{2k+1}$. The value $(H_o(x+1)-H_o(x))-(H_e(x+1)-H_e(x))=\frac{1}{2x+3}-\frac{1}{2x+2}$, which means that the sequence decreases. $H_o(x)\ge H_e(x)$ since $H_o(1)=1$ and $H_e(1)=\frac{1}{2}$, $\frac{1}{3}>\frac{1}{4}$, and so on. Since the sequence decreases and is greater than $0$, it converges.

Answer 1

As noticed your proof looks fine. As an alternative we have that:

$$\sum_{k=0}^x\frac{1}{2k+1}-\sum_{k=1}^x\frac{1}{2k} =\frac{1}{2x+1}+\sum_{k=0}^{x-1}\frac{1}{2k+1}-\sum_{k=1}^x\frac{1}{2k}=$$

$$=\frac{1}{2x+1}+\sum_{k=1}^{x}\frac{1}{2k-1}-\sum_{k=1}^x\frac{1}{2k}=\frac{1}{2x+1}+\sum_{k=1}^{x}\frac{1}{2k(2k-1)}$$

which clearly converges by comparison test.