# Urgend Geometric series question

arildno said:
Correct! That series have you seen before in this thread; what does it sum up to?

Hi it sums up to

$(-1)^{n} x^{2n}$

/Fred

arildno
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No it does not; that is a TERM in the series.
What does the whole series sum up to?

arildno
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Mathman23 said:
$\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1) + (-1)x^{2} + (-1)x^{4} + (-1)x^{6} + (-1)x^{8}$

/Fred
OOPS! This is wrong!
When j is even, $(-1)^{j}=1$ wheras if j is odd, we have $(-1)^{j}=-1$

Use this info.

Hi the series summes up to $$\frac{1}{1-x^2}$$

/Fred

arildno said:
No it does not; that is a TERM in the series.
What does the whole series sum up to?

Hi

Then

$\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}$

Because j is even.

/Fred

arildno said:
OOPS! This is wrong!
When j is even, $(-1)^{j}=1$ wheras if j is odd, we have $(-1)^{j}=-1$

Use this info.

arildno
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No, it sums up to 1/(1+x^{2}).

arildno
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Mathman23 said:
Hi

Then

$\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}$

Because j is even.

/Fred
No, j alternates being even and odd.

Mathman23 said:
Hi

Then

$\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}$

Because j is even.

/Fred
No, j is not only even! If that would be the case, your expression would be $x^{4} + x^{8} + x^{12}+...$. Go back and check again.

arildno said:
No, it sums up to 1/(1+x^{2}).

Then

$\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}$ Sums up to

$$\frac{1}{(1+x^{2})}$$

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arildno said:
No, it sums up to 1/(1+x^{2}).

Then

$\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}$ Sums up to 

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First of all, $\sum_{j=0}^{\infty}(-1)^{j}x^{2j} \neq x^{2} + x^{4} + x^{6} + x^{8}$, it is $\sum_{j=0}^{\infty}(-1)^{j}x^{2j}=1-x^{2} + x^{4} - x^{6} + x^{8}...$.

arildno
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No. You MUST learn to read properly and stop writing sloppy and nonsensical maths.

We have:
$$(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}$$

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And second; you know also this: $\frac{1}{1-q}=1 + q + q^2 + q^3 +...$.

Can you see the connection now?

I strongly suggest you to over the entire problem again and - as arildno is saying - read it properly.

arildno said:
No. You MUST learn to read properly and stop writing sloppy and nonsensical maths.

We have:
$$(-1)^{n}x^{2n}\sum_{j=0}^{\infty}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}$$

Then
$$\sum_{j=0}^{\infty}(-1)^{j}x^{2j}=1-x^{2} + x^{4} - x^{6} + x^{8} +\cdots + (-1)^{n}x^{2n} = \sum_{j=0}^{\infty}\frac{(-1)^{n}x^{2n}}{1+x^{2}}$$

arildno said:
We have:
$$(-1)^{n}x^{2n}\sum_{j=0}^{\infty}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}$$
Sorry arildno, but that doesn't make sense.

arildno