Vtyjkyuk

Does the series beginaligns_n(x)=sum^n_k=0frace^-kx1+kx,;xin [1,2]endalign
converge uniformly?

MY TRIAL:

I thought, by intuition, that $s_n(x)$ does not converge uniformly, which implies that $s_n(x)$ is not uniformly Cauchy i.e., $exists,epsilon_0>0$ s.t. $forall,kin BbbN, ,exists,n_k,m_kgeq k,;exists,x_kin[1,2]$ s.t. beginalignleft|s_n_k -s_m_kright|geq epsilon_0.endalign

Let $kin BbbN$ be given. Take $n_k=k,,m_k=k+1,,x_k=1+frac1k+1$ and . Then,
beginalignleft|s_n_k -s_m_kright|&=left|sum^n_k_p=0frace^-px_k1+px_k-sum^m_k_p=0frace^-px_k1+px_kright|\&=left|sum^k_p=0frace^-px_k1+px_k-sum^k+1_p=0frace^-px_k1+px_kright|\&=left|frace^-(k+1)left(1+frac1k+1right)1+(k+1)left(1+frac1k+1right)right||\&=left|frace^-k+23+kright|endalign
From here, I don't know what to do. Any help please? Anyway, I don't even know if my guess was right. It could be that the series converges uniformly.

Hint : Use Weierstrass-M test and the fact that $|frace^-k x1+k x |leq frace^-k1+k$ for all $x in [1,2]$

Edit : $ 1+k leq 1+ k x => k leq k x => 1 leq x$ and this is true since $x in [1,2]$, so the denominator is the smallest when $x=1$.

$ e^-k x leq e^-k $ => applying log to both sides of the inequality gives $ -k x leq -k$ => dividing by $-k$ and flipping the inequality (division by negative number) to get $x geq 1$ and this is true since $x in [1,2]$

By Weierstrass-M test and the fact of $sum limits_k=1^infty frace^-k1+k$ convergence we get that the original sum is uniformly convergent when $x in [1,2]$.