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Problem

Ratio test

Part 1: Consider the following infinite series. Determine whether each infinite series converges or diverges.

$ \sum_{n=1}^\infty\frac{n}{2^n} $
$ \sum_{n=1}^\infty\frac{2^n}{n} $

Part 2: Consider the following infinite series. Determine the radius of convergence and the interval of convergence.

$ \sum_{n=1}^\infty\frac{4^n}{n}(x-3)^n $

Solution

Part 1

Applying the ratio test for each series.

Series 1

$ \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| $
$ \lim_{n\to\infty} \left| \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}\right| $
$ \lim_{n\to\infty} \left| \frac{n+1}{2n} \right| = \frac{1}{2} $

Since this limit is less than 1, the series converges.

Series 2

$ \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| $
$ \lim_{n\to\infty} \left| \frac{\frac{2^{n+1}}{n+1}}{\frac{2^n}{n}} \right| $
$ \lim_{n\to\infty} \left| \frac{2n}{n+1} \right| = 2 $

Since this limit is greater than 1, the series diverges.

Part 2

In order for the series to converge, the limit must be less than 1.

$ \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 $

For the series

$ \sum_{n=1}^\infty\frac{4^n}{n}(x-3)^n $,
$ \lim_{n\to\infty} \left| \frac{4^{n+1}}{n+1} {(x-3)}^{n+1} \cdot \frac{n}{4^n {(x-3)}^{n}} \right| < 1 $
$ \lim_{n\to\infty} \left| \frac{4n}{n+1}(x-3) \right| < 1 $
$ \lim_{n\to\infty} \left| \frac{4n}{n+1}\right| |x-3| < 1 $
$ 4|x-3| < 1 $
$ |x-3| < \frac{1}{4} $

At this stage, we find that the radius of convergence is $ R = \frac{1}{4} $.

To solve for the interval of convergence, just solve for $ x $.

$ \frac{-1}{4} < x-3 < \frac{1}{4} $
$ \frac{11}{4} < x < \frac{13}{4} $.

This is as far as the ratio test allows us to conclude. To determine if each endpoint converges or not, we must apply other tests.

When $ x = 2.75 $, the series becomes

$ \sum_{n=1}^\infty\frac{4^n}{n}{\left(\frac{-1}{4}\right)}^{n} $
$ \sum_{n=1}^\infty\frac{(-1)^n}{n} $

By the alternating series test, this series converges.

When $ x = 3.25 $, the series becomes

$ \sum_{n=1}^\infty\frac{4^n}{n}{\left(\frac{1}{4}\right)}^{n} $
$ \sum_{n=1}^\infty\frac{1}{n} $

This is the Harmonic series, and it was shown here that it diverges.

Therefore, the radius of convergence for

$ \sum_{n=1}^\infty\frac{4^n}{n}(x-3)^n $

is actually $ \frac{11}{4} \le x < \frac{13}{4} $.

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