‎Is the functional sequence ‎$f_n(x)$ ‎uniformly ‎convergent‎ ‎on ‎$‎[1, 2]‎$‎?

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Consider the functional sequence ‎‎\begin{align*}‎ ‎f_n(x) = ‎\sqrt{n+1} - \sqrt{n+2} + \sum_{k=1}^n ‎\frac{1}{2\sqrt{k+x+2}} - ‎‎‎‎\frac{1}{2\sqrt{k+x+1}}, (1\leq x\leq 2, n=1,2,3,...). ‎‎\end{align*}‎‎‎ Is ‎$‎f_n(x)‎$ ‎uniformly ‎convergent ‎on ‎‎$‎[1, 2]‎$?‎ ‎

‎ I‎ ‎know ‎that ‎‎$f(x) =‎ \displaystyle{\lim_{n\to\infty}}f_n(x) = \sum_{k=1}^\infty‎\frac{1}{2‎\sqrt{k+x+2}‎} - ‎‎‎‎‎\frac{1}{2‎\sqrt{k+x+1}‎}‎ = ‎‎‎‎\frac{-1}{2‎\sqrt{x+2}‎}‎‎‎$,‎ but I do not know how to use the supremm test‎. ‎Please ‎guide ‎me?‎ (Let ‎$‎M_n = sup|f_n(x) - f(x)|‎$‎, so ‎$‎f_n‎‎\rightarrow ‎f‎$ ‎is ‎uniformly ‎convergent ‎on ‎‎$‎E‎$ ‎if ‎and ‎only ‎if ‎‎$‎M_n‎‎\rightarrow ‎0‎$ ‎as ‎‎$‎n‎‎\rightarrow ‎\infty‎$‎)‎.

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For $x\in[1,2]$, $$ \begin{align}‎ ‎f_n(x) &=\sqrt{n+1}-\sqrt{n+2}+\sum_{k=1}^n\left(\frac1{2\sqrt{k+x+2}}-\frac1{2\sqrt{k+x+1}}\right)\\ &=-\underbrace{\frac1{\sqrt{n+1}+\sqrt{n+2}}}_{\le\frac1{2\sqrt{n+1}}}+\underbrace{\frac1{2\sqrt{x+n+2}}}_{\le\frac1{2\sqrt{n+3}}}-\frac1{2\sqrt{x+2}}\\ ‎‎\end{align} $$ Can you finish from here?