Question

Find the interval of convergence of the series (n(2x-5)^n)/(n+2) from n=0 to infinity?

Answer 1

@zepdrix

Answer 2

did you do:
\[\lim_{n \rightarrow \infty} |\frac{(n+1)(2x-5)^{n+1}}{n+2+1} \cdot \frac{n+2}{n(2x-5)^n}|\]

Answer 3

\[\sum_{n=0}^{\infty}\frac{n(2x-5)^n}{n+2} \]
consider your \(\large a_n=\frac{n}{n+2}\)
So, the sum converges for \(|2x-5|<1\), and you can isolate for |x| if you want.

Answer 4

So when I took that limit, I got 1 times (2x-5) which is still 1 for the answer. But how did you get abs(2x-5)<1?

Answer 5

the \((2x-5)^{n+1}\) will cancel with the \((2x-5)^n\) to give \(2x-5\) in the numerator. You also have in the numerator: (n+1)(n+2) which gives \(n^2+3n+2\) when multplied out, and the denominator has (n)(n+3) = \(n^2+3n\)

Thus: \[ \left| \frac{(2x-5)(n^2+3n+2)}{n^2+3n}\right|\]

The \(n^2\) term dominates, so as \(n\rightarrow \infty\), the n terms go to 1, and your left with \( |2x-5|\), and by the ratio test, this converges if this quantity is < 1

Answer 6

You just keep the absolute value along the ride as you find the limit. Don't get rid of them.

Answer 7

Thank you so much!

Answer 8

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