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## Problem

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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