Ratio test gives me $\displaystyle \lim_{n\to\infty}\frac{\sqrt{2(n+1)}-3\sqrt{1+2(n+1)}+3\sqrt{2+2(n+1)}-\sqrt{3+2(n+1)}}{\sqrt{2n}-3\sqrt{1+2n}+3\sqrt{2+2n}-\sqrt{3+2n}}=1$, I'm not sure how to proceed.
Mathematica gives
-Sqrt[2] Sqrt[n]-Sqrt[1+2 n]+
4 ((-1+2 Sqrt[2]) Zeta[-(1/2)]+
(-1)^(2 n) Sqrt[2] Zeta[-(1/2),1/2+n,IncludeSingularTerm->False]
+(-1)^(1+2 n) Sqrt[2] Zeta[-(1/2),1+n,IncludeSingularTerm->False])
And is numerically converging to $-1.52041\ldots$.
Let $S_n = 4\sum\limits_{k=1}^{2n} (-1)^k \sqrt{k} - \sqrt{2n}-\sqrt{2n+1}$ be the finite sum at hand. We are going to show
Unlike its exact value, the existence of $S_\infty$ can be justified using elementary methods.
Let $a_k$ be the coefficient of $\sqrt{k}$ in above expression of $S_n$. Notice
$$\begin{align}\sum_{k=1}^{2n+1} a_k t^k &= 4\sum_{k=1}^{2n}(-1)^k t^k - t^{2n}(1+t) = -4t\frac{1-t^{2n}}{1+t} + (1-t^{2n})(1+t) -(1+t)\\ &= -1-t + ((1+t)^2 - 4t)\frac{1-t^{2n}}{1+t} = -1-t + (1-2t+t^2)\sum_{k=1}^{2n}(-1)^{k-1} t^{k-1}\end{align}$$ We can rewrite $S_n$ to have the form of an alternating sum: $$S_n = -1 + \sum_{k=1}^{2n} (-1)^{k}b_k\quad\text{ where }\quad b_k = 2\sqrt{k} - \sqrt{k-1} - \sqrt{k+1}$$
Notice as $k$ increases, $$\begin{align}b_k &= (\sqrt{k}-\sqrt{k-1}) - (\sqrt{k+1}-\sqrt{k})\\ &= \frac{1}{\sqrt{k}+\sqrt{k-1}} - \frac{1}{\sqrt{k+1}+\sqrt{k}} = \frac{\sqrt{k+1}-\sqrt{k-1}}{(\sqrt{k}+\sqrt{k-1})(\sqrt{k+1}+\sqrt{k})}\\ &= \frac{2}{(\sqrt{k+1}+\sqrt{k-1})(\sqrt{k}+\sqrt{k-1})(\sqrt{k+1}+\sqrt{k})} \end{align} $$ monotonically decreases to zero. By alternating series test, $-1 + \sum\limits_{k=1}^\infty (-1)^k b_k$ is a converging series. Since $S_n$ is a sub-sequence of its partial sums, $S_n$ converges.
To evaluate $S_\infty$, let $T_n = \sum\limits_{k=1}^{2n}(-1)^k\sqrt{k}$. We have $S_n = 4T_n - \sqrt{2n} - \sqrt{2n+1}$ and $$T_n = \left(\sum_{k=1,\text{even}}^{2n} - \sum_{k=1,\text{odd}}^{2n}\right)\sqrt{k} = \left(2\sum_{k=1,\text{even}}^{2n} - \sum_{k=1}^{2n}\right)\sqrt{k} = \sqrt{8}\sum_{k=1}^n \sqrt{k} - \sum_{k=1}^{2n}\sqrt{k} $$ It is known that the sum over $\sqrt{k}$ has following asymptotic expansion (e.g. see this answer) $$\sum_{k=1}^n k^{1/2} \asymp \frac23 n^{3/2} + \frac12 n^{1/2} + K + \frac{1}{24}n^{-1/2} + \cdots\quad\text{ where } K = \zeta\left(-\frac12\right)$$
Substitute this expansion into expression of $T_n$, we find $$\begin{align} S_n &= 4\sqrt{8}\left(\frac23 n^{3/2} + \frac12 n^{1/2} + K\right) - 4\left(\frac23 (2n)^{3/2} + \frac12 (2n)^{1/2} + K\right) - 2\sqrt{2n} + O(n^{-1/2})\\ &= 4(\sqrt{8}-1)K + O(n^{-1/2}) \end{align} $$ As a result, $S_\infty = \lim\limits_{n\to\infty} S_n = 4(\sqrt{8}-1)\zeta\left(-\frac12\right) \approx -1.520419250438736...$
I will show that the limit exists.
If $s_n =\left(4\sum_{i=1}^{2n}(-1)^i\sqrt{i}\right)-\sqrt{2n}-\sqrt{2n+1} $ then, using $\sqrt{1+x} =1+\frac{x}{2}-\frac{x^2}{8} + O(x^3) $ with the constant in the big-oh satisfying $0 < c < \frac1{16}$,
$\begin{array}\\ s_{n+1}-s_n &=4(\sqrt{2n+2}-\sqrt{2n+1})-(\sqrt{2n+2}+\sqrt{2n+3})+(\sqrt{2n}+\sqrt{2n+1})\\ &=3\sqrt{2n+2}-3\sqrt{2n+1}-\sqrt{2n+3}+\sqrt{2n}\\ &=\sqrt{2n}(3\sqrt{1+1/n}-3\sqrt{1+1/(2n)})-\sqrt{1+3/(2n)}+1)\\ &=\sqrt{2n}(3(1+\frac{1}{2n}-\frac{1}{8n^2} + O(\frac1{n^3})) -3(1+\frac{1}{4n}-\frac{1}{32n^2} + O(\frac1{n^3})) -(1+\frac{3}{4n}-\frac{9}{32n^2} + O(\frac1{n^3}))+1)\\ &=\sqrt{2n}(\frac{1}{n}(\frac32 -\frac34-\frac34) -\frac{1}{n^2}(-\frac38+\frac{3}{32}+\frac{9}{32})+ O(\frac1{n^3})))\\ &=O(\frac1{n^{5/2}})\\ \end{array} $
and the sum of this converges (because of the bounded coefficients of the $O(\frac1{n^3})$ terms).
