# Finding value of converging series

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?

• 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. – Clement C. Mar 8 '16 at 5:15
• Sorry, it is $(-4)^n$. I'll edit it to show this. – dibdub Mar 8 '16 at 5:16
• Ah, I see what I did. Got it, thanks! – dibdub Mar 8 '16 at 5:20
• Then your answer is correct: this is $\frac{1}{4x-19}$ indeed, for $(-4)^n(x-5)^n$. – Clement C. Mar 8 '16 at 5:20

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.