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Consider the geometric series $\; \displaystyle \sum_{n=0}^{\infty} \, (-1)^{n}(x+6)^{n} .$

The open interval of convergence is ?

Find the sum of the series on this interval ?

Formula for calculating sum of the geometric series is:

$$\frac{1}{1 - r}$$

Where sum must look as:

$\sum_{n=0}^{\infty} r^n\;\;$

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Anyhow, to find sum:

I rewrote the equation as:

$ \sum ((-x-6)^2)^n $

Then sum is:

$ \frac{1}{1 - (-x-6)^2} = \frac{1}{-x^2 - 12x - 35}$

But it's not right, how come I should use as my r $(-x-6)$ and not $(-x-6)^2$ ??? I thought that r is whatever inside the parenthesis when it's in the power of "n".

Consider the other infinite series:

$ \sum_{n=0}^{\infty} \left(\frac{1}{ 4}\right)^{\!n/2}$

To use sum formula:

$\frac{a}{1 - r}$

Then you rewrite it as $((\frac{1}{4})^{\frac{1}{2}})^n$

Then to use formula , your r would be $(\frac{1}{4})^{1/2} $ not just $\frac{1}{4}$

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  • $\begingroup$ You would have more chance of being answered if the quality of your question was better. $\endgroup$ – hlapointe Oct 22 '15 at 2:58
  • $\begingroup$ 1. $|x+6| < 1 \iff -1 < x+6 < 1 \iff -7 < x < -5 $ 2. Why have you introduced the square? What about terms like $ -(x+6)$? $\endgroup$ – stochasticboy321 Oct 22 '15 at 3:10
  • $\begingroup$ @stochasticboy321 Can't you see how I rewrote the equation? And essentially, that's what my r is, r is whatever the x inside is (x)^n. In my case, inside ^n is (-x-6)^2 $\endgroup$ – Jack Oct 22 '15 at 3:20
  • $\begingroup$ You just have to consider $\sum_{n=0}^\infty a^n$ with $a=-x-6$. Apply the formula for GP. $\endgroup$ – Claude Leibovici Oct 22 '15 at 3:22
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    $\begingroup$ Mate, look at your original problem. Is there a square in it? Sure, you rewrote the series, but is the new series the same as the old series? Think carefully, aren't you dropping terms from the original series when you square each term? $\endgroup$ – stochasticboy321 Oct 22 '15 at 18:35
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Since $\sum_{n \geq 0} y^n = \frac{1}{1-y}$, for $y = -(x+6)$ we have $$\sum_{n \geq 0} (-1)^n (x+6)^n = \color{blue}{\sum_{n \geq 0} y^n = \frac{1}{1-y} } = \frac{1}{x+7}.$$ To find where $\sum_{n \geq 0} (-1)^n (x+6)^n$ converges, you can use that the blue equality holds for $|y| < 1$.

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To find the sum, devide this series into two series- that which has even value of 'n' and that which has odd. Then, add the individual results.

Lets assume $\ x+6= y$ for convenience.

$ \sum_{n=0}^{\infty} \, (-1)^{n}(y)^{n}= (y^0 + y^2 + y^4 +...) - (y^1 +y^3 +y^5 +...)$

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  • $\begingroup$ But you can't assume that because you don't just have $y^n$ you have $(y^2)^n$ $\endgroup$ – Jack Oct 22 '15 at 17:55
  • $\begingroup$ I do not get what you are saying. Please explain why I cannot assume y = x+6. $\endgroup$ – Prem kumar Oct 27 '15 at 8:25
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You were absolutely right, except it appears that you had confused exponential product rule for exponential power of a product rule.

So your rewritten equation would look like:

$\sum_{n=0}^{\infty} \, (-x-6)^{n}$

Using sum formula:

$\frac{1}{1 - (-x - 6)} $ = $\frac{1}{x+7}$

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