Question

HELP PLEASE!

Answer 1

I dont understand why -1 is a closed interval for #2.

Answer 2

@zepdrix @518nad @mathmale

Answer 3

Using Ratio Test to Solve (Please tell me If I am doing this correctly or not...)
\[\sum_{n=1}^{\infty}\frac{ x^n }{ 9n-1 }\rightarrow \lim_{n \rightarrow \infty}\left| \frac{ x^{n+1} }{ 9(n+1)-1 } \times \frac{ 9n-1 }{ x^n }\right|\]

Answer 4

\[\left| x \right|\lim_{n \rightarrow \infty}\left| \frac{ 9n-1 }{ 9n+8 } \right|\]
\[\left| x \right| < 1 \rightarrow (-1,1)\]

Answer 5

But the answer says the Interval of Convergence is [-1,1)

Answer 6

plug in x=-1 into the original \[ \[\sum_{\infty}^{n=1}\]\frac{ (-1)^n }{ 9n-1 }\] it's an alternation series
apply the alternating series test

Answer 7

Wouldn't that Converge. The graph would look like this.