Determine the limit of sequence $a_n=1+\frac{1}{3}+\frac{1}{5}+ ...+\frac{1}{2n-1}-\ln(\sqrt{n})$ if it exists else explain why it doesn't exist.
My try: I tried to use thm that if $a_n$ is bounded and monotonic then it is convergent. $$a_{n+1}-a_n=\frac{1}{2n+1}-\ln(\sqrt{n+1})+\ln(\sqrt{n})=\frac{1}{2n+1}+\frac{1}{2}\ln(\frac{n}{n+1})=\frac{1}{2n+1}+\frac{1}{2}\ln(1-\frac{1}{n+1})>\frac{1}{2n+1}-\frac{1}{2n+2}>0$$ But don't know to bound it above.
In a elementary fashion we have $$\begin{eqnarray*}\log\left(\frac{n+1}{n}\right)&=&\int_{n}^{n+1}\frac{dx}{x}=\int_{0}^{1}\frac{dx}{n+x}=\int_{-1/2}^{1/2}\frac{dx}{n+\frac{1}{2}+x}\\&=&\int_{0}^{1/2}\left(\frac{1}{n+\frac{1}{2}+x}+\frac{1}{n+\frac{1}{2}-x}\right)\,dx\\&=&\int_{0}^{1}\frac{dz}{\left(n+\frac{1}{2}\right)-\frac{z^2}{4n+2}}\end{eqnarray*}$$ so for any $n\geq 1$ $$ \log\left(\frac{n+1}{n}\right)-\frac{1}{\left(n+\frac{1}{2}\right)}=\frac{1}{4}\int_{0}^{1}\frac{z^2}{\left(n+\frac{1}{2}\right)^3-\frac{z^2}{4}\left(n+\frac{1}{2}\right)}\,dz$$ is bounded by $\frac{1}{12}\cdot\frac{1}{n\left(n+\frac{1}{2}\right)(n+1)}$ and it is decreasing towards zero as $n\to +\infty$.
This essentially is a naive implementation of the Hermite-Hadamard inequality.
Since $\log\left(\frac{n+1}{n}\right)$ and $\frac{1}{\left(n+\frac{1}{2}\right)}$ are pretty close (with the former being a telescopic term) and $\sum_{n\geq 1}\frac{1}{n\left(n+\frac{1}{2}\right)(n+1)}$ is finite (equal to $\frac{3-4\log 2}{6}$) your sequence is convergent.
The Euler-Mascheroni constant $\gamma$ is exactly defined as $\lim_{n\to +\infty}\left(H_n-\log n\right)$. From $$ H_n = \log(n) + \gamma + O\left(\frac{1}{n}\right) $$ we have $$ H_{2n}-\frac{1}{2}H_n-\frac{1}{2}\log(n) = \underbrace{\color{red}{\frac{\gamma}{2}+\log 2}}_{\approx 0.981755}+O\left(\frac{1}{n}\right).$$
Long story short, your limit is just a bit less than one.