MHB Finding the difference of two series?

[shamieh]

$$A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2}$$...

So I know this is the series $$\sum^{\infty}_{n = 1} \frac{(-1)^n}{n^2}$$

Then I'm given

$$S =1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2}$$...

And I know that this is obviously the series $$\sum^{\infty}_{n = 1} \frac{1}{n^2}$$

Question: Given that $$S = \frac{\pi^2}{6}$$ find the sum A given above. (Hint: consider S - A and express it in terms of S)

So I think I know the answer to this but I'm not sure and might need some guidance...But if both of these series are convergent since the alternating series is convergent by using the absolute value test and the original 1/n^2 is convergent by p series, if we were to take S - A wouldn't we get every other term remaining , while every other term canceled ? Like for example

$$S - A$$

$$S = 1 + 1/2^2 $$
$$A = 1 - 1/2^2 $$

So the second terms would cancel right? and the first term sum would be 2.. but every odd$$^{th} $$term seems to remain as it will double since both signs are positive... How would I do this?

 

[chisigma]

$$A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2}$$...

So I know this is the series $$\sum^{\infty}_{n = 1} \frac{(-1)^n}{n^2}$$

Then I'm given

$$S =1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2}$$...

And I know that this is obviously the series $$\sum^{\infty}_{n = 1} \frac{1}{n^2}$$

Question: Given that $$S = \frac{\pi^2}{6}$$ find the sum A given above. (Hint: consider S - A and express it in terms of S)

So I think I know the answer to this but I'm not sure and might need some guidance...But if both of these series are convergent since the alternating series is convergent by using the absolute value test and the original 1/n^2 is convergent by p series, if we were to take S - A wouldn't we get every other term remaining , while every other term canceled ? Like for example

$$S - A$$

$$S = 1 + 1/2^2 $$
$$A = 1 - 1/2^2 $$

So the second terms would cancel right? and the first term sum would be 2.. but every odd$$^{th} $$term seems to remain as it will double since both signs are positive... How would I do this?

It is easy to see that $\displaystyle A = S - 2\ \frac{S}{2^{2}} = \frac{S}{2}$...

Kind regards

$\chi$ $\sigma$

 

[Prove It]

$$A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2}$$...

So I know this is the series $$\sum^{\infty}_{n = 1} \frac{(-1)^n}{n^2}$$

Then I'm given

$$S =1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2}$$...

And I know that this is obviously the series $$\sum^{\infty}_{n = 1} \frac{1}{n^2}$$

Question: Given that $$S = \frac{\pi^2}{6}$$ find the sum A given above. (Hint: consider S - A and express it in terms of S)

So I think I know the answer to this but I'm not sure and might need some guidance...But if both of these series are convergent since the alternating series is convergent by using the absolute value test and the original 1/n^2 is convergent by p series, if we were to take S - A wouldn't we get every other term remaining , while every other term canceled ? Like for example

$$S - A$$

$$S = 1 + 1/2^2 $$
$$A = 1 - 1/2^2 $$

So the second terms would cancel right? and the first term sum would be 2.. but every odd$$^{th} $$term seems to remain as it will double since both signs are positive... How would I do this?

$\displaystyle \begin{align*} S &= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \dots \\ S &= \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \\ S &= \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \dots + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \\ S &= A + 2\left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \right) \\ S &= A + 2 \left[ \frac{1}{ \left( 2 \cdot 1 \right) ^2} + \frac{1}{ \left( 2 \cdot 2 \right) ^2 } + \frac{1}{ \left( 2 \cdot 3 \right) ^2 } + \dots \right] \\ S &= A + 2 \left[ \frac{1}{4} \left( \frac{1}{1^2} \right) + \frac{1}{4} \left( \frac{1}{2^2} \right) + \frac{1}{4} \left( \frac{1}{3^2} \right) + \dots \right] \\ S &= A + \frac{1}{2} \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots \right) \\ S &= A + \frac{1}{2}S \\ A &= \frac{1}{2} S \\ A &= \frac{1}{2} \left( \frac{\pi ^2}{6} \right) \\ A &= \frac{ \pi ^2 }{12} \end{align*}$