# Application of squeeze theorem

I don't really know how the Squeeze theorem works and I tried applying it to solve this limit:

$$\lim_{n\to\infty}\text{ } \frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(n+n)^2}$$ So $$\frac{n}{(n+n)^2}\leq \frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(n+n)^2} \leq \frac{n}{n^2}$$ Then:

$$\lim_{n\to\infty}\text{ } \frac{n}{n^2}=0$$

$$\lim_{n\to\infty}\text{ } \frac{n}{(n+n)^2}=0$$

Therefore: $$\lim_{n\to\infty}\text{ } \frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(n+n)^2}=0$$

Is this the correct way?

You should have $$n \to \infty$$ not $$x \to \infty$$.
You can afford to be sloppier with the lower bound ($$0$$ will work), and your upper bound will be much easier to verify if you use $$\dfrac{n+1}{n^2}$$.
what you have is: $$L=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{(n+k)^2}$$ and if we make the substitution $$u=n+k$$ we can obtain: $$L=\lim_{n\to\infty}\sum_{u=n}^{2n}\frac{1}{u}^2$$ which we can rewrite as: $$L=\lim_{n\to\infty}\left[\sum_{u=0}^{2n}\frac{1}{u^2}-\sum_{u=0}^{n-1}\frac{1}{u^2}\right]$$ and when $$n\to\infty$$ these two summations are equal, so it is equal to $$0$$