$\sum_{n=1}^{\infty} u_n^2=0$ $\Rightarrow$ $u_n=0 \ \forall \ n\in \mathbb{N}$ Let, $<u_n>$ be a real sequence and given that $$\sum_{n=1}^{\infty} u_n^2=0$$
Prove that $u_n=0 \ \forall \ n\in \mathbb{N}$
Attempt
$u_n^2\geq 0 \ \forall \ n\geq 1$
Since, $$\sum_{n=1}^{\infty} u_n^2=0.$$Please prove this result.
Sum of infinitely many non negative terms is 0 if all the terms are 0
 A: If the series sums zero this means that the sequence of partial sums,
for an arbitrary $\varepsilon>0$ and a suitable $n$ satisfies,  $$u^2_1+u^2_2+...+u^2_n\le\varepsilon,$$
then each $u^2_k\le\varepsilon$, so $u_k=0$.
A: Assume there exists at least one nonzero term, say $u_k$. Then, for $n\geq k$, the partial sum $S_n\geq u_k^2>0$. Therefore:
$$\lim_{n \to \infty}S_n=\sum_{n=1}^\infty u_n^2>0$$
which is a contradiction.
So we conclude every term of the sequence must be 0.
A: We can prove a more general result:

If $\{x_n\}_{n=0}^{\infty}$ is a nonnegative sequence and
$\sum_{n=0}^{\infty}x_n=0$, then $x_n=0$ for all $n\in\mathbb{N}$.

Let $\{s_k\}_{k=0}^{\infty}$ be the partial sums of the series $\sum_{n=0}^{\infty}x_n$ and $R_k=\sum_{n=k+1}^{\infty}x_n$ the corresponding remainder. We're given that $\sum_{n=0}^{\infty}x_n=0$, so for every $k\geq 0$, we can write
\begin{align*}
0 &= \sum_{n=0}^{\infty}x_n\\
&= \sum_{n=0}^{k}x_n+\sum_{n=k+1}^{\infty}x_n\\
&= s_k + R_k\\
\iff s_k &= -R_k
\end{align*}
The terms $x_n$ in each sum are nonnegative, so $s_k\geq 0 $ and $R_k\geq 0$, implying that $s_k\geq 0$ and $-R_k\leq 0$. This gives $0\leq s_k=-R_k\leq 0$, so $s_k$ must be identically $0$. We deduce that for all $k\geq 0$,
\begin{align*}
x_k &= s_k-s_{k-1}\\
&= 0-0\\
&= 0
\end{align*}
The result you want is the special case $x_n=a_{n}^2$.
