$$\dfrac{1}{n} \sum_{i=1}^n \dfrac{1}{i^2}$$ is the given sequence. I know that sums of converging sequences are convergent also, so $\dfrac{1}{i^2}=0$ and thus the sum is also convergent. I also know that products of convergent sequences are also convergent, if both converge. $\dfrac{1}{n}=0$ so I have $0*0=0$ which shows that the sequence given above is convergent and its limit is $0$.
Correct or horribly wrong?
On
We have that
$$0\le \dfrac{1}{n} \sum_{i=1}^n \dfrac{1}{i^2}\le \frac 1n\frac{\pi^2}{6}$$
then by squeeze theorem
$$ \dfrac{1}{n} \sum_{i=1}^n \dfrac{1}{i^2}\to 0$$
On
You have two sequences.
One is the sequence $$\frac {1}{n}$$ and the other one is $$ \sum_{i=1}^n \dfrac{1}{i^2}$$
The first one approaches $0$ and the second on approaches $ \frac {\pi ^2 }{6}$
Thus the product $$\dfrac{1}{n} \sum_{i=1}^n \dfrac{1}{i^2}$$ will approach $0\times \frac {\pi ^2 }{6} =0$
Just wrong, not horribly wrong,
The sum of a fixed number of convergent sequences is convergent, but in your case you have that $n$ above the sum sign. Using the integral test (for instance), you can prove that the sum $\sum_{i=1}^n\frac1{i^2}$ converges. Since $\lim_{n\to\infty}\frac1n=0$, the limit of your sequence is indeed $0$.