How to prove that the series $\sum\limits_{n=1}^\infty \left(\left(n+\frac12\right)\ln\left(1+\frac1n\right) - 1 \right)$ converges

Find an equivalent of the general term. We'll have to expand the log at order $3$: \begin{align}\Bigl(n+\frac12\Bigr)\ln\Bigl(1+\frac1n\Bigr)-1&=\Bigl(n+\frac12\Bigr)\Bigl(\frac1n-\frac 1{2n^2}+\frac1{3n^3}+o\Bigl(\frac1{n^3}\Bigr)\Bigr)-1\\&=1-\frac1{2n}+\frac1{3n^2}+o\Bigl(\frac1{n^2}\Bigr)+\frac1{2n}-\frac 1{4n^2}+\frac1{6n^3}+o\Bigl(\frac1{n^3}\Bigr)-1\\ &=\frac 1{12n^2}+o\Bigl(\frac1{n^2}\Bigr) \end{align} Thus $\;u_n\sim_\infty\dfrac 1{12n^2}$, which converges.

By L'Hopital's Rule, one has \begin{eqnarray} &&\lim_{n\to\infty}\frac{(n+\frac12)\ln(1+\frac1n)-1}{\frac1{n^2}}\\ &=&\lim_{x\to0}\frac{(\frac1x+\frac12)\ln(1+x)-1}{x^2}\\ &=&\lim_{x\to0}\frac{(x+2)\ln(1+x)-2x}{2x^3}\\ &=&\frac1{12}. \end{eqnarray} Noting that $\sum_{n=1}^\infty\frac1{n^2}$ converges and so $\sum_{n=1}^\infty\big[(n+\frac12)\ln(1+\frac1n)-1\big]$ converges.

Update: Without using L'Hopital's Rule. Note $$ \ln(1+x)\le x-\frac{x^2}{2}+\frac13x^3, x>0. $$ Using this, one has \begin{eqnarray} 0\le(n+\frac12)\ln(1+\frac1n)-1&\le&(n+\frac12)(\frac1n-\frac{1}{2n^2}+\frac1{3n^3})-1\\ &=&\frac{1}{12}\frac1{n^2}+\frac{1}{6n^3} \end{eqnarray} So $\sum_{n=1}^\infty\frac1{n^2}$ converges.

We can prove that the sum converges by proving that the limit of the partial sums converges. We can do that by simply solving the sum!

Define: $$ S(N) = \sum_{n = 1}^{N-1} \left[ \left(n + \frac{1}{2}\right)\ln\left(1 + \frac{1}{n}\right) - 1\right]\, . $$ (We define the upper limit as $N-1$ rather than $N$ because it makes the math "prettier" later.) I can rearrange this as follows: \begin{align} S(N) &= \sum_{n = 1}^{N-1} \left[ \left(n + \frac{1}{2}\right)\ln\left(\frac{n+1}{n}\right)\right] - (N-1)\\ &= \sum_{n = 1}^{N-1} \left[ \left(n + 1 + \frac{1}{2}\right)\ln\left(n+1\right) - \left(n + \frac{1}{2}\right)\ln\left(n\right) - \ln(n+1) \right] - (N-1)\\ &= \sum_{n = 1}^{N-1} \left[ \left(n + 1 + \frac{1}{2}\right)\ln\left(n+1\right) - \left(n + \frac{1}{2}\right)\ln\left(n\right) \right] - \ln\left(N!\right)- (N-1) \end{align} Note that the argument of the remaining sum can be written as $$ \sum_{n = 1}^{N-1} \left[ \left(n + 1 + \frac{1}{2}\right)\ln\left(n+1\right) - \left(n + \frac{1}{2}\right)\ln\left(n\right) \right] = \sum_{n = 1}^{N-1}\left[f(n+1) - f(n)\right]\, , $$ where $f(x) = \left(x + \frac{1}{2}\right)\ln(x)$. Thus this sum telescopes, leaving: $$ \sum_{n = 1}^{N-1} \left[ \left(n + 1 + \frac{1}{2}\right)\ln\left(n+1\right) - \left(n + \frac{1}{2}\right)\ln\left(n\right) \right] = f(N) - f(1) = \left(N+\frac{1}{2}\right)\ln(N) $$ Putting it all together: $$ S(N) = \left(N+\frac{1}{2}\right)\ln(N)- \ln\left(N!\right)- (N-1) $$ We want to take the limit of $S(N)$ as $N\rightarrow\infty$, so we invoke the Stirling approximation for the factorial: \begin{align} \lim_{N\rightarrow\infty}S(N)&= \lim_{N\rightarrow\infty}\left\{ \left(N+\frac{1}{2}\right)\ln(N)- \ln\left(N^N e^{-N} \sqrt{2\pi N}\,\right)- (N-1)\right\}\\ &= 1 - \ln\left(\sqrt{2\pi}\right)\approx 0.08106 \end{align}

(In the last step above, all the terms inside the limit involving $N$ cancel each other out, so we're just left with the constant.)

Numerically evaluating the sum (in Mathematica) gives the same result.