Question

series question!!

Answer 1

Use the ratio test for finding the radius of convergence. You can read about it here:
http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

So divide a^n+1 by a^n, like so

\[\Large L=\lim_{n \rightarrow \infty }\left| \frac{ (x+8)^{n+1} }{ 5^{n+1} }*\frac{ 5^n }{ (x+8)^n }\right|\]

Answer 2

Then simplify (I'm not showing nearly all the steps, you'll have to do it yourself on paper) \[\Large L=\lim_{n \rightarrow \infty }\left| \frac{ (x+8) }{ 5 }\right|\]
\[\Large L=(x+8)\lim_{n \rightarrow \infty }\left| \frac{ 1 }{ 5 }\right|\]
\[\Large L=(x+8)\frac{ 1 }{ 5 }\]

Answer 3

If L < 1, the series converges. I forgot the absolute value around x+8, so solve this for x.
\[\Large \frac{ 1 }{ 5 }\left| x+8 \right|<1\]

Answer 4

When you get the solutions for the equation above,
let's call these solutions, \(\color{#000000}{ \displaystyle x=a }\) and \(\color{#000000}{ \displaystyle x=b }\),

THEN, you will have to plug in \(\color{#000000}{ \displaystyle x=a }\) and \(\color{#000000}{ \displaystyle x=b }\)
into the series and see if the series converges at
these values of x.

(Well, in this case, for either \(\color{#000000}{ \displaystyle x=a }\) and \(\color{#000000}{ \displaystyle x=b }\) series will diverge. Either because you get 1+1+1+1+1+1.... or because you get 1+(-1)+1+(-1)+1+(-1)....