I ran into this limit while evaluating an integral.
$$\lim_{N\to\infty}\left(2\sqrt{N+1}\;-\;\sum_{n=1}^N\frac1{\sqrt{n}}\right)$$
I know this converges, but I don’t know how to approach this, mostly because of the sum inside the limit. Using Desmos I found out that the limit is approximately 1.46
We can also use Dirichlet eta-function $\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}$ to define this limit.
First of all we see that eta-function is defined (the series converges) for $s>0$ (Dirichlet convergence test). Next, we use the relation between eta-function and zeta-function for $s>1$: $$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}=\sum_{n=1}^\infty\frac{1}{n^s}-2\sum_{n=1}^\infty\frac{1}{(2n)^s}=\big(1-2^{1-s}\big)\zeta(s)$$ Using this relation we can define zeta-function for $s\in(0;1)$ as $\zeta(s)=\frac{\eta(s)}{1-2^{1-s}}$ We can also present eta-function in the form $$\lim_{N\to\infty}\sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n^s}=\lim_{N\to\infty}\bigg(\sum_{n=1}^{2N}\frac{1}{n^s}-2\sum_{n=1}^N\frac{1}{(2n)^s}\bigg)=\lim_{N\to\infty}\bigg(\sum_{n=1}^{N}\frac{1-2^{1-s}}{n^s}+\sum_{n=N+1}^{2N}\frac{1}{n^s}\bigg)$$ The second term can be presented as the Riemann sum at $N\to\infty$ $$I_2(N)=\sum_{n=N+1}^{2N}\frac{1}{n^s}=\frac{N}{N^S}\frac{1}{N}\sum_{n=N+1}^{2N}\frac{1}{\Big(\frac{n}{N}\Big)^s}\to N^{1-s}\int_1^2\frac{dx}{x^s}=N^{1-s}\frac{2^{1-s}-1}{1-s}$$ $$\eta(s)=\lim_{N\to\infty}\bigg(\sum_{n=1}^{N}\frac{1-2^{1-s}}{n^s}-N^{1-s}\frac{1-2^{1-s}}{1-s}\bigg)$$ $$\zeta(s)=\lim_{N\to\infty}\bigg(\sum_{n=1}^{N}\frac{1}{n^s}-\frac{N^{1-s}}{1-s}\bigg)\,;\,\,\, s\in(0;1)$$
Our initial limit $$\lim_{N\to\infty}\left(2\sqrt{N+1}\;-\;\sum_{n=1}^N\frac1{\sqrt{n}}\right)=\lim_{N\to\infty}\left(2\sqrt{N}\;-\;\sum_{n=1}^N\frac1{\sqrt{n}}\right)=-\zeta\big(\frac{1}{2}\big)=1.46035...$$ Zeta-function
$$A_N=2\sqrt{N+1}\;-\;\sum_{n=1}^N\frac1{\sqrt{n}}=2\sqrt{N+1}-H_N^{\left(\frac{1}{2}\right)}$$ Using the binomial expansion and the asymptotics of generalized harmobic numbers $$A_n=-\zeta \left(\frac{1}{2}\right)+\frac 1{2\sqrt N}\Bigg[1-\frac{5}{12 N}+\frac{1}{4 N^2}+O\left(\frac{1}{N^3}\right)\Bigg]$$
The answer to the question you posed is, yes. Also, your answer looks very close to $3/2$. I recommend looking at old posts that are similar to your own question, such as this one: Closed form of $\lim\limits_{n\to\infty}\left(\int_0^{n}\frac{{\rm d}k}{\sqrt{k}}-\sum_{k=1}^n\frac1{\sqrt k}\right)$
A lot of the time, limits of the form $$\sum_{k=1}^n f(k) - \int_c^n f(k)$$ will exist, but won't have names. Sure, we have Euler-Mascheroni, but that's more the exception than the rule!
(Also, a name won't necessarily give you more info about the constant; a numerical approximation, on the other hand, is usually good enough!)
