# Science:Infinite Series Module/Units/Unit 2/2.4 The Ratio Test/2.4.04 A Simple Ratio Test Example

## Question

Using only the ratio test, determine whether or not the recursive sequence

align}"): {\displaystyle \begin{align} a_1 = 1, \qquad a_{k+1} = \frac{0.1 + \cos(k)}{\sqrt{k}}a_k, \quad k = 1, 2, 3, \ldots \end{align}

converges or diverges.

## Complete Solution

Applying the ratio test yields

align}"): {\displaystyle \begin{align} \lim_{k \rightarrow \infty} \Big| \frac{a_{k+1}}{a_k} \Big| &= \lim_{k \rightarrow \infty} \Bigg| \frac{\frac{0.1 + \cos(k)}{\sqrt{k}}a_k}{a_k} \Bigg| && (1) \\ &= \lim_{k \rightarrow \infty} \Bigg| \frac{0.1 + \cos(k)}{\sqrt{k}} \Bigg| && (2)\\ \end{align}

But

{\displaystyle {\begin{aligned}0&\leq \lim _{k\rightarrow \infty }{\Bigg |}{\frac {0.1+\cos(k)}{\sqrt {k}}}{\Bigg |}\leq \lim _{k\rightarrow \infty }{\frac {1.1}{\sqrt {k}}}&&(3)\\0&\leq \lim _{k\rightarrow \infty }{\Bigg |}{\frac {0.1+\cos(k)}{\sqrt {k}}}{\Bigg |}\leq 0&&(4)\\\end{aligned}}}

Therefore,

{\displaystyle {\begin{aligned}\lim _{k\rightarrow \infty }{\Bigg |}{\frac {0.1+\cos(k)}{\sqrt {k}}}{\Bigg |}=0&&(5)\\\end{aligned}}}

Since the limit equals ${\displaystyle 0}$, the ratio test tells us that the series converges.

## Explanation of Each Step

### Step (1)

To apply the ratio test, we must evaluate the limit

{\displaystyle {\begin{aligned}\lim _{k\rightarrow \infty }{\Big |}{\frac {a_{k+1}}{a_{k}}}{\Big |}\end{aligned}}}

In our problem, we can use

{\displaystyle {\begin{aligned}a_{k+1}={\frac {0.1+\cos(k)}{\sqrt {k}}}a_{k}\end{aligned}}}

and substitute this into our limit.

### Step (2)

In Step (2), we only cancel the ${\displaystyle a_{k}}$ in the numerator and denominator.

### Step (3)

First observe that

{\displaystyle {\begin{aligned}0\leq |0.1+\cos(k)|\leq 1.1\end{aligned}}}

Dividing everything by the square root of ${\displaystyle k}$ we obtain

{\displaystyle {\begin{aligned}0\leq {\Big |}{\frac {0.1+\cos(k)}{\sqrt {k}}}{\Big |}\leq {\frac {1.1}{\sqrt {k}}}\end{aligned}}}

### Step (4)

In Step (4) we only evaluate the limit:

{\displaystyle {\begin{aligned}\lim _{k\rightarrow \infty }{\frac {1.1}{\sqrt {k}}},\end{aligned}}}

which equals zero because the numerator is a constant and the denominator goes to infinity.

### Step (5)

In Step (5) we apply the Squeeze Theorem.

## Potential Challenge Areas

### Getting Started

Because the question asks us to apply the ratio test, we know that we will start our solution by using the formula

{\displaystyle {\begin{aligned}\lim _{k\rightarrow \infty }{\Big |}{\frac {a_{k+1}}{a_{k}}}{\Big |}.\end{aligned}}}

### Recursive Formula

Most problems involving convergence tests don't involve recursive formulas. But with the ratio test, we apply

{\displaystyle {\begin{aligned}\lim _{k\rightarrow \infty }{\Big |}{\frac {a_{k+1}}{a_{k}}}{\Big |}\end{aligned}}}

and use the given recursion equation for ${\displaystyle a_{k+1}}$. In our case, our recursion equation is

{\displaystyle {\begin{aligned}a_{k+1}={\frac {0.1+\cos(k)}{\sqrt {k}}}a_{k}\end{aligned}}}

which we substitute into the numerator, allowing us to cancel the ${\displaystyle a_{k}}$ in the numerator and denominator. This trick is a bit harder to apply for the other convergence tests.