We have $\lim_{n\rightarrow \infty}(a_n+b_n)=a+b$
if $\lim_{n\rightarrow \infty}a_n=a$ and $\lim_{n\rightarrow \infty}b_n=b$
Why is this wrong:
$\lim_{n\rightarrow \infty}1=\lim_{n\rightarrow \infty}\left(\frac{1}{n}+ \frac{1}{n}+ \cdots + \frac{1}{n}\right)=n\lim_{n\rightarrow \infty}\frac{1}{n}=0$?
You have, as a proven theorem, that if $\lim_{n\to\infty} a_n$ exists and is equal to $a$, and $\lim_{n\to\infty} b_n$ exists and is equal to $b$, then $$ \lim_{n\to\infty} (a_n+b_n) = a+b $$ Analagous theorems are true for finite sums of any size, e.g., $\lim_{n\to\infty} (a_n+b_n+c_n + d_n) = a+b+c+d$.
You don't have a theorem that says
If for all $i\in\Bbb{N}: \lim_{n\to\infty} x_n^{(i)}$ exists and is equal to $x^{(i)}$, then $$\lim_{n\to\infty}\left( \sum_{i=1}^\infty x_n^{(i)} \right) = \sum_{i=1}^\infty x^{(i)} $$ which in words would say that you can always interchange the order of summing and taking the limit, even in infinite sums.
If you did have that as a theorem then you could use it to validate your "proof" that $1=0$. In fact, your observation is a proof by contradiction that sometimes interchanging the sum and limit operations gives different answers.
Congratulations on discovering this important "you must not assume this" fact on your own!