Consider the remainder given by
$$R_n(x)=\int^{0}_{x}\frac{(x-t)^{n-1}}{(n-1)!}f^{n}(t)dt,$$
Prove that $$|R_n|\leq{n}B|a_n||x|,$$
where $B=\max\limits_{t\in[0,x]}\left|\dfrac{(x-t)^{n-1}}{1+t}\right|(1+t)^{-1/2}$
This is part (d) of a multi part question which is supposed to prove that the Taylor series for $f(x)=\sqrt{1+x}$ converges to $f$ for all $x\in(0,1)$.
For part (a) I used the ratio test on $a_k = \dfrac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2) \cdots (\frac{1}{2}-k+1)}{k!}$ to show radius of convergence is $1$.
I'm just not sure how to proceed so far I have only seen Cauchy's version of the remainder which has no integrals, so I'm not sure how to find $|R_n|$.
I'm not sure how to integrate this since there are three variables $x,t$ and $n$.
Any help would be greatly appreciated.
Thank you.