= Solution
Use the deterministic weights $a_{n,k}=\sqrt{k}/n$. Their squared sum and maximum are
$$
\sum_{k=1}^na_{n,k}^2=\frac{n+1}{2n}\longrightarrow\frac12,
\qquad \max_{k\leq n}|a_{n,k}|=n^{-1/2}\longrightarrow0.
$$
To apply the <quadratic expansion of a cumulant-generating function> rigorously, write $\log M_X(u)=u^2/2+r(u)$, with $r(u)=o(u^2)$. Fix $t\ne0$ and let
$$
\eta_n=\sup_{0<|u|\leq|t|/\sqrt n}\frac{|r(u)|}{u^2}\longrightarrow0.
$$
All arguments $t\sqrt{k}/n$ lie in this shrinking interval. Consequently
$$
\left|\sum_{k=1}^nr\!\left(\frac{t\sqrt{k}}n\right)\right|
\leq\eta_n\frac{t^2}{n^2}\sum_{k=1}^nk
=\eta_n\frac{t^2(n+1)}{2n}\longrightarrow0.
$$
Taking logs of the <moment-generating function> and using <independence of random variables> now gives
$$
\log M_{T_n}(t)=\frac{t^2}{2n^2}\sum_{k=1}^nk+o(1)
\longrightarrow\frac{t^2}{4}.
$$
The case $t=0$ is immediate. Since $e^{t^2/4}$ is the <moment-generating function> of a <normal distribution> of variance $1/2$, the <convergence theorem for moment-generating functions> gives
$$
\boxed{T_n\xrightarrow dN\!\left(0,\frac12\right).}
$$
This is a <weighted central limit theorem with negligible maximum weight>. Its limit variance also agrees with the exact value $\operatorname{Var}(T_n)=(n+1)/(2n)$ from <variance additivity for independent random variables>.
Back to article page
