Ratio Test

## **Problem**

**Part 1:** Determine whether each infinite series converges or diverges:

(
1
)

∑

# n
1

∞

n

2
n

(
2
)

∑

# n
1

∞

2
n

n

{\displaystyle \begin{align}
(1) \, \sum\_{n=1}^\infty\frac{n}{2^n} \\[5 pt]
(2) \, \sum\_{n=1}^\infty\frac{2^n}{n}
\end{align} }

**Part 2:** Determine the radius of convergence and the interval of convergence of the infinite series

∑

# n
1

∞

4
n

n

(
x
−
3

)
n

.

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}(x-3)^n. }

## **Solution**

**Part 1**

Apply the ratio test for each series.

**Series 1**

lim

n
→
∞

a

n
1

a
n

{\displaystyle \lim\_{n\to\infty} \left| \frac{a\_{n+1}}{a\_n} \right| }

lim

n
→
∞

n
1

2

n
1

n

2
n

{\displaystyle \lim\_{n\to\infty} \left| \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}\right| }

lim

n
→
∞

n
1

2
n

=

1
2

{\displaystyle \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
→
∞

a

n
1

a
n

{\displaystyle \lim\_{n\to\infty} \left| \frac{a\_{n+1}}{a\_n} \right| }

lim

n
→
∞

2

n
1

n
1

2
n

n

{\displaystyle \lim\_{n\to\infty} \left| \frac{\frac{2^{n+1}}{n+1}}{\frac{2^n}{n}} \right| }

lim

n
→
∞

2
n

n
1

=
2

= 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
→
∞

a

n
1

a
n

<
1

{\displaystyle \lim\_{n\to\infty} \left| \frac{a\_{n+1}}{a\_n} \right| < 1 }

For the series

∑

# n
1

∞

4
n

n

(
x
−
3

)
n

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}(x-3)^n }

,

lim

n
→
∞

4

n
1

n
1

(
x
−
3
)

n
1

⋅

n

4
n

(
x
−
3
)

n

<
1

{\displaystyle \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
→
∞

4
n

n
1

(
x
−
3
)

<
1

{\displaystyle \lim\_{n\to\infty} \left| \frac{4n}{n+1}(x-3) \right| < 1 }

lim

n
→
∞

4
n

n
1

x
−
3

<
1

4

x
−
3

<
1

{\displaystyle 4|x-3| < 1 }

x
−
3

<

1
4

.

{\displaystyle |x-3| < \frac{1}{4}. }

At this stage, we find that the radius of convergence is

# R

1
4

{\displaystyle R = \frac{1}{4} }

. To solve for the interval of convergence, determine the interval for

x

{\displaystyle x}

.

−
1

4

<
x
−
3
<

1
4

{\displaystyle \frac{-1}{4} < x-3 < \frac{1}{4} }

11
4

<
x
<

13
4

.

{\displaystyle \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

{\displaystyle x = 2.75 }

, the series becomes

∑

# n
1

∞

4
n

n

(

−
1

4

)

n

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}{\left(\frac{-1}{4}\right)}^{n} }

∑

# n
1

∞

(
−
1

)
n

n

.

{\displaystyle \sum\_{n=1}^\infty\frac{(-1)^n}{n}. }

By the alternating series test, this series converges.

When

# x
3.25

{\displaystyle x = 3.25 }

, the series becomes

∑

# n
1

∞

4
n

n

(

1
4

)

n

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}{\left(\frac{1}{4}\right)}^{n} }

∑

# n
1

∞

1
n

.

{\displaystyle \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

∑

# n
1

∞

4
n

n

(
x
−
3

)
n

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}(x-3)^n }

 is actually

11
4

≤
x
<

13
4

{\displaystyle \frac{11}{4} \le x < \frac{13}{4} }

.