# [SOLVED] Series (p-series)

• December 9th 2009, 10:54 AM
nameck
[SOLVED] Series (p-series)
if my an = 1 / (n+1)^2
can i use p series test to confirm that the series will converge..
where p = 2..
• December 9th 2009, 11:18 AM
osolage
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.
• December 9th 2009, 11:19 AM
nameck
ok..so.. what test should i use to know whethere it is convergence or divergence?
• December 9th 2009, 11:31 AM
Plato
Quote:

Originally Posted by nameck
if my an = 1 / (n+1)^2
can i use p series test to confirm that the series will converge..where p = 2..

Quote:

Originally Posted by osolage
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)

Well of course it works. One just has to understand the process.
$\sum\limits_{n = 1}^\infty {\frac{1}{{\left( {n + 1} \right)^2 }}} = \sum\limits_{k = 2}^\infty {\frac{1}{{\left( k \right)^2 }}}$
The series on the right is a convergent p-series.

Of notice that $\frac{1}{(n+1)^2}\le \frac{1}{n^2}$ use comparison test.
• December 9th 2009, 11:36 AM
osolage
polar to cartesian
deleted mssg