# Science:Infinite Series Module/Units/Unit 2/2.4 The Ratio Test/2.4.05 Ratio Test Example with Exponent

## Question

Using only the ratio test, determine whether or not the series

align}"): {\displaystyle \begin{align} \sum_{k=1}^{\infty} \frac{k}{5^k} \end{align}

converges, diverges, or yields no conclusion.

## 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{(k+1) /5^{k+1}}{k/5^k} \Bigg| && (1) \\ &= \lim_{k \rightarrow \infty} \Bigg| \frac{k+1}{k} \cdot \frac{5^k}{5^{k+1}} \Bigg| && (2) \\ &= \lim_{k \rightarrow \infty} \Bigg( \Bigg| \frac{k+1}{k} \Bigg| \cdot \Bigg| \frac{5^k}{5^{k+1}} \Bigg| \Bigg) && (3) \\ &= \lim_{k \rightarrow \infty} \Bigg| \frac{k+1}{k} \Bigg| \cdot \lim_{k \rightarrow \infty}\Bigg| \frac{5^k}{5^{k+1}} \Bigg| && (4) \\ &= 1 \cdot \lim_{k \rightarrow \infty}\Bigg| \frac{5^k}{5^{k+1}} \Bigg| && (5) \\ &= \lim_{k \rightarrow \infty}\Bigg| \frac{5^k}{5 \cdot 5^k} \Bigg| && (6) \\ &= \lim_{k \rightarrow \infty}\Bigg| \frac{1}{5 } \Bigg| \\ &= \frac{1}{5}. \end{align}

Since the limit equals ${\displaystyle 1/5}$, 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 have

{\displaystyle {\begin{aligned}a_{k}={\frac {k}{5^{k}}},\quad \mathrm {and} \quad a_{k+1}={\frac {k+1}{5^{k+1}}},\end{aligned}}}

and we substitute them into our limit.

### Step (2)

In Step (2), we use a little algebraic manipulation to make things easier to look at

{\displaystyle {\begin{aligned}{\frac {(k+1)/5^{k+1}}{k/5^{k}}}&={\frac {(k+1)/5^{k+1}}{k/5^{k}}}\cdot {\frac {5^{k+1}}{5^{k+1}}}&&(2.1)\\&={\frac {(k+1)/1}{k/5^{k}}}\cdot {\frac {1}{5^{k+1}}}&&(2.2)\\&={\frac {k+1}{k/5^{k}}}\cdot {\frac {5^{k}}{5^{k}}}\cdot {\frac {1}{5^{k+1}}}&&(2.3)\\&={\frac {k+1}{k}}\cdot {\frac {5^{k}}{1}}\cdot {\frac {1}{5^{k+1}}}&&(2.4)\\\end{aligned}}}

### Step (3)

Step (3) uses a property of absolute values. Recall that for real numbers ${\displaystyle a}$ and ${\displaystyle b}$,

{\displaystyle {\begin{aligned}|a\cdot b|=|a|\cdot |b|.\end{aligned}}}

### Step (4)

Step (4) uses a property of limits values. Recall that for functions ${\displaystyle f(k)}$ and ${\displaystyle g(k)}$, ${\displaystyle k\in \mathbb {R} }$ that

{\displaystyle {\begin{aligned}\lim _{k\rightarrow \infty }{\Big (}f(k)\cdot g(k){\Big )}=\lim _{k\rightarrow \infty }f(k)\cdot \lim _{k\rightarrow \infty }g(k)\end{aligned}}}

### Step (5)

Here we evaluate a limit:

{\displaystyle {\begin{aligned}\lim _{k\rightarrow \infty }{\Bigg |}{\frac {k+1}{k}}{\Bigg |}=\lim _{k\rightarrow \infty }{\Bigg |}{\frac {1+k^{-1}}{1}}{\Bigg |}=1.\end{aligned}}}

### Step (6)

Some algebraic manipulation helps us see how we can simplify our problem. Recall that, using laws of exponentials, that for real numbers ${\displaystyle x,a,b}$, that

{\displaystyle {\begin{aligned}x^{a+b}=x^{a}\cdot x^{b},\end{aligned}}}

so

{\displaystyle {\begin{aligned}5^{k+1}=5\cdot 5^{k}.\end{aligned}}}