Prove that $\sum_{k=1}^{n} a_k \left(1-\frac{k^2}{n^2}\right) \to 1$ 
Problem : Given $a_n$ which satisfies $\displaystyle\sum a_n = 1$, prove that
$$\lim_{n\to\infty}\sum_{k=1}^na_k\left(1-\frac{k^2}{n^2}\right) = 1.$$


My Attempt
From $\displaystyle\sum a_n = 1$, I changed this problem to show $\displaystyle\lim\sum\frac{a_kk^2}{n^2}\to 0$.
From given condition, I know that $a_n \to 0$ and it means that $a_n$ is bounded.
So, for arbitrary large $n$, $|a_n|<1$, and I tried to bound an original sum
$$\left|\sum\frac{a_kk^2}{n^2}\right|\le \frac{1}{n^2}\left|\sum a_kk^2\right|\le \frac{1}{n^2}\sum k^2$$
but I failed because the right term goes to infinity.
Thanks for help.
 A: The standard approach to tacking abelian-type results like this is to consider $s_n = \sum_{k=1}^{n} a_k$ and write the sum in terms of $(s_n)$ using summation by parts:
\begin{align*}
\sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right)
&= \sum_{k=1}^{n} (s_k - s_{k-1}) \left( 1 - \frac{k^2}{n^2} \right) \\
&= \sum_{k=1}^{n} s_k \left( 1 - \frac{k^2}{n^2} \right) - \sum_{k=1}^{n} s_{k-1} \left( 1 - \frac{k^2}{n^2} \right) \\
&= \sum_{k=1}^{n-1} s_k \left( 1 - \frac{k^2}{n^2} \right) - \sum_{k=1}^{n-1} s_{k} \left( 1 - \frac{(k+1)^2}{n^2} \right) \\
&= \sum_{k=1}^{n-1} s_k \frac{2k+1}{n^2}.
\end{align*}
Then by using $\sum_{k=1}^{n-1} \frac{2k+1}{n^2} = 1 - \frac{1}{n^2}$, the difference $d_n$ between the sum $\sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right)$ and the limit candidate $1$ is bounded by
\begin{align*}
d_n := \left| \sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right) - 1 \right|
&= \left| \sum_{k=1}^{n-1} (s_k - 1) \frac{2k+1}{n^2} + \frac{1}{n^2} \right| \\
&\leq \sum_{k=1}^{n-1} \left| s_k - 1 \right| \frac{2k+1}{n^2} + \frac{1}{n^2}.
\end{align*}
Now, for an arbitrary $\varepsilon > 0$, choose $N$ such that $|s_k - \varepsilon| < \varepsilon$ whenever $k \geq N$. Then for $n > N$, by splitting the last sum into two parts, one over $k < N$ and the other over $k \geq N$, $d_n$ is further bounded by
\begin{align*}
d_n
&\leq \sum_{k=1}^{N-1} \left| s_k - 1 \right| \frac{2k+1}{n^2}
+ \sum_{k=N}^{n-1} \left| s_k - 1 \right| \frac{2k+1}{n^2}
+ \frac{1}{n^2} \\
&\leq \left( \max_{k < N} |s_k - 1| \right) \left( \sum_{k=1}^{N-1} \frac{2k+1}{n^2} \right)
+ \varepsilon \left( \sum_{k=N}^{n-1} \frac{2k+1}{n^2} \right)
+ \frac{1}{n^2} \\
&\leq \left( \max_{k < N} |s_k - 1| \right) \frac{N^2}{n^2}
+ \varepsilon
+ \frac{1}{n^2}.
\end{align*}
Letting limsup as $n\to\infty$, this bound reduces to
\begin{align*}
\limsup_{n\to\infty} d_n
&\leq \varepsilon
\end{align*}
However, since the left-hand side does not depend on $\varepsilon$, we may let $\varepsilon \to 0^+$ to obtain
$$ \limsup_{n\to\infty} \left| \sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right) - 1 \right| = 0. $$
Therefore the desired assertion follows.
