Answer
Converges to $1$
Work Step by Step
Consider $f(x)=\dfrac{1}{x^2}$. This is positive, continuous and decreasing for $x \geq 1$
Now, take the integral test to find the convergence and divergence for the sequence.
We have $\int_1^\infty \dfrac{1}{x^2} dx= \lim\limits_{a \to \infty} \int_1^a \dfrac{1}{x^2} dx$
or, $ \lim\limits_{a \to \infty} [\dfrac{-1}{x}]_1^a=\lim\limits_{a \to \infty} [\dfrac{-1}{a}+1]=1$
Hence, the sequence $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ converges to $1$
