if my an = 1 / (n+1)^2
can i use p series test to confirm that the series will converge..
where p = 2..
no it wont work. the p series test only works when the denominator is n raised to a power 1/n or 1/n^2 or 1/n^(3/2)
even when you indirectly use the p series test with direct comparison or limit comparison an must still be in the form of 1/n
the integral test works the integral 1/(x+1)^2 ,x,0,infinity converges thus the series converges ... the sum however is not found that way.
Well of course it works. One just has to understand the process.
$\displaystyle \sum\limits_{n = 1}^\infty {\frac{1}{{\left( {n + 1} \right)^2 }}} = \sum\limits_{k = 2}^\infty {\frac{1}{{\left( k \right)^2 }}} $
The series on the right is a convergent p-series.
Of notice that $\displaystyle \frac{1}{(n+1)^2}\le \frac{1}{n^2}$ use comparison test.