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`sum_(n=1)^oo (1/n^2-1/n^3)` Determine the convergence or divergence of the series.

`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.

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