Uniform convergence in $\mathbb{R}^2$ $a$, $b$ are $2$ points in $\mathbb{R}^{2},\rho_{n}(t)\,:\,[0,1]\to\mathbb{R}^{2}$ is a sequence of continuously differentiable constant speed curves with $\|\rho_n'(t)\|=L_n$ for all $t$ from $0$ to $1$ and $\rho_n(0)=a,\,\rho_{n}(1)=b,\forall n$. Suppose that $\lim_{n\to\infty}L_n=\| b-a\|$. I need to show that $\rho_n$ converges uniformly to $\rho(t)=a+t(b-a)$ for $t\in[0,1]$.
So intuitively it's converging to the straight line, for a proof I was thinking about a theorem for uniform convergence of series that says if the series of derivatives converge uniformly to some some function $k$ and the original series converge point wise on some point, then the original series converges uniformly to some $g$ who's derivative is exactly the function $k$.  So I think this fits all the criterion except that this is from $\mathbb{R}$ to $\mathbb{R}^{2}$ and I'm not sure if the function extends in this case.  ie it's not the derivative here that converges uniformly but rather the arclength so to speak.  Is there a similar result for higher dimensions?  Thanks!
 A: Since you are also interested in the high dimensional case and since there is no difference in the argument, let me assume that $\rho_n:[0,1]\to\mathbb{R}^d$ is a sequence of curves in $\mathbb{R}^d$ for some $d\ge 2$, which satisfies all of your other assumptions.

It suffices to consider the case $a\ne b$. Denote $L=\|b-a\|$ and $\mathbf{u}=\frac{b-a}{L}$. To show $\rho_n$ uniformly converges to $\rho$, it suffices to show that for every $n\ge 1$ and every $t\in[0,1]$, 
$$\|\rho_n(t)-\rho(t)\|^2\le L_n^2-L^2 .\tag{1}$$
Given $n\ge 1$ and $t\in[0,1]$, there exist $\mathbf{v}\in\mathbb{R}^d$ and $x,y\in\mathbb{R}$, such that $\|\mathbf{v}\|=1$, $\mathbf{v}\bot\mathbf{u}$ and $\rho_n(t)-\rho(t)=x\mathbf{u}+y\mathbf{v}$. 
If $x\ge 0$, then from $\rho_n(t)-a=(x+tL)\mathbf{u}+y\mathbf{v}$ we know that
$$\|\rho_n(t)-\rho(t)\|^2=x^2+y^2\le (x+tL)^2+y^2-t^2L^2=\|\rho_n(t)-a\|^2-t^2L^2.\tag{2}$$
Since 
$$\|\rho_n(t)-a\|=\|\int_0^t \rho_n'(s)d s\|\le tL_n,\tag{3}$$ 
$(1)$ follows in this case.
If $x< 0$, then similar to $(2)$ and $(3)$, we have $\|\rho_n(t)-\rho(t)\|^2\le \|\rho_n(t)-b\|^2-(1-t)^2L^2$ and $\|\rho_n(t)-b\|\le(1-t)L_n$. $(1)$ also follows.

Remark added: In the argument above, we didn't use the assumption that $\rho_n$ has constant speed. Actually we proved the following statement. 

Proposition: Given $d\ge 2$ and $a,b\in\Bbb R^d$, denote $\rho(t)=a+t(b-a)$, $\forall t\in[0,1]$. Let $\{L_n\}$ be a sequence of real numbers with $\lim\limits_{n\to\infty}L_n=\|b-a\|$. For every $n\ge 1$, let $\rho_n:[0,1]\to \Bbb R^d$ be a continuously differentiable curve satisfying that $\rho_n(0)=a$, $\rho_n(1)=b$ and $\|\rho_n'(t)\|\le L_n$, $\forall t\in[0,1]$(this implies $L_n\ge \|b-a\|$). Then $\rho_n$ converges uniformly to $\rho$.

