`sum_(n=1)^oo(1/n^2-1/n^3)`

Apply the series sum/difference rule:

`=sum_(n=1)^oo1/n^2-sum_(n=1)^oo1/n^3`

Observe that both the series are p-series of the form`sum_(n=1)^oo1/n^p`

Recall that the p-series test is applicable for the series of the form `sum_(n=1)^oo1/n^p` ,where `p>0`  

If `p>1` , then the p-series converges

If `0<p<=1` , then the p-series diverges

Now, both the series have `p>1`

As per the p-series test , both the series converge and so their sum/difference will also converge.

Hence the series `sum_(n=1)^oo(1/n^2-1/n^3)` converges.