Question # Determine the convergence or divergence of the series. sum_{n=1}^infty(frac{1}{n^2}-frac{1}{n^3})

Series
ANSWERED Determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty(\frac{1}{n^2}-\frac{1}{n^3})$$ 2020-11-30
P-series test:
$$\sum_{n=1}^\infty\frac{1}{n^p}$$
If $$p>1$$ series is convergent
If $$p\leq1$$ series is divergent
Series sum or difference rule:
If $$\sum_{n=1}^\infty a_n$$ and $$\sum_{n=1}^\infty b_n$$ are both convergent then
$$\sum_{n=1}^\infty(a_n\pm b_n)=\sum_{n=1}^\infty a_n\pm\sum_{n=1}^\infty b_n$$
And sum of two convergent series is also convergent
Given that $$\sum_{n=1}^\infty\frac{1}{n^p}$$
Let $$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{1}{n^2}$$ and $$\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty\frac{1}{n^3}$$
Using p-series test both $$\sum a_n$$ and $$\sum b_n$$ are convergent so using series sum or difference rule $$\sum_{n=1}^\infty(\frac{1}{n^2}-\frac{1}{n^3})=\sum_{n=1}^\infty\frac{1}{n^2}-\sum_{n=1}^\infty\frac{1}{n^3}$$
Since $$\sum_{n=1}^\infty a_n=\frac{1}{n^2}$$ and $$\sum_{n=1}^\infty b_n=\frac{1}{n^3}$$ both are convergent. Hence $$\sum_{n=1}^\infty(\frac{1}{n^2}-\frac{1}{n^3})$$ is also convergent