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I cannot for the life of me figure out what I am doing wrong with this. The question posed to me was to find the values of x for which the following series converges:

$$ \sum_{i=0}^\infty (-4)^n*(x-5)^n $$

I calculated that x must be in the range $ \frac{19}{4} < x < \frac{21}{4} $, which I believe is correct.

However, when I try to calculate the value to which the series converges when x is within this range, my algebra is off by a negative sign. The correct answer is $ \frac{1}{4x-21}$, but I keep getting $\frac{1}{4x-19}$

My algebra is as follows: $$ r = (-4)(x-5) = (20-4x) $$ $$a=1$$ $$ \sum_{i=0}^\infty (-4)^n*(x-5)^n = \frac{a}{1-r} = \frac{1}{1-(20-4x)}$$ $$= \frac{1}{4x-19} $$

It's got to be a simple thing I'm overlooking, can anyone spot it?

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    $\begingroup$ Is it $-4^n(x-5)^n$, or $(-4)^n(x-5)^n$? This does make quite a big difference, in particular to compute your $r$: this flips the sign... And it looks like you picked the wrong one. $\endgroup$ – Clement C. Mar 8 '16 at 5:15
  • $\begingroup$ Sorry, it is $(-4)^n$. I'll edit it to show this. $\endgroup$ – dibdub Mar 8 '16 at 5:16
  • $\begingroup$ Ah, I see what I did. Got it, thanks! $\endgroup$ – dibdub Mar 8 '16 at 5:20
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    $\begingroup$ Then your answer is correct: this is $\frac{1}{4x-19}$ indeed, for $(-4)^n(x-5)^n$. $\endgroup$ – Clement C. Mar 8 '16 at 5:20
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See $-1$ is a number noT related to n as it isnt $(-4)^n$ know the difference ! Then do your algebra you will get it ie $\frac{a}{1-r}=\frac{1}{1-(4x-20)}=\frac{1}{(-4x+21)}$ and then taking negative sign down we get the answer.

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  • $\begingroup$ Yup, thats exactly what I did. I actually did the algebra correct but when I checked the answer on Wolfram I forgot to include the parenthesis, which gave me the wrong answer. Thanks! $\endgroup$ – dibdub Mar 8 '16 at 5:21
  • $\begingroup$ Hope you got it now $\endgroup$ – Archis Welankar Mar 8 '16 at 5:25

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