Alternating series remainder theorem

[hgaon001]

Use the theorem to approximate the sum from n=1 to infinity of ((-1)^n)/(n^2) with an error <_ .0001

i know that a_n+1 has to be less than a_n but what do i do after i prove that?

 

[galactus]

$\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}$

In the event you're interested, the series consists of the negative of the sum of the squares of the reciprocals of the ODDS, and the sum of the squares of the reciprocals of the EVENS.

$\displaystyle -1+\frac{1}{4}-\frac{1}{9}+\frac{1}{16}-\frac{1}{25}+...................$

$\displaystyle \left(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+......\right)-\left(1+\frac{1}{9}+\frac{1}{25}+............\right)$

$\displaystyle \sum_{n=1}^{\infty}\frac{-1}{(2n-1)^{2}}=\frac{-{\pi}^{2}}{8}$

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n)^{2}}=\frac{{\pi}^{2}}{24}$

$\displaystyle \frac{{\pi}^{2}}{24}-\frac{{\pi}^{2}}{8}=\frac{-{\pi}^{2}}{12}$

 

[Deleted member 4993]

Use the theorem to approximate the sum from n=1 to infinity of ((-1)^n)/(n^2) with an error <_ .0001

i know that a_n+1 has to be less than a_n but what do i do after i prove that?

Please tell us - what does

Alternating series remainder theorem

state.

Then show us how you tried to apply that to the series.