Ratio Test for Series: Statement, Solved Examples

The ratio test for series is useful to check a series is convergent or divergent. Here we consider the limit of the quotient of (n+1)-th and n-th terms of the series, and depending on this limit is greater or less than 1 we decide its convergence. In this article, we study the ratio test of a series along with some solved examples.

Ratio Test Statement

Let $\displaystyle \sum_{n=1}^\infty a_n$ be a series of positive real numbers. Suppose that

ℓ = lim_n→∞ $\dfrac{a_{n+1}}{a_n}$.

Then we have the following.

• If ℓ<1, then the series ∑a_n converges.
• If ℓ>1, then ∑a_n diverges.
• If ℓ=1, we cannot conclude.

Ratio Test Solved Examples

$\boxed{\color{blue}\textbf{Q}1:}$ Test the convergence of $2+\dfrac{2^2}{2}+\dfrac{2^3}{3}+\cdots$.

$\boxed{\color{red}\textbf{Answer:}}$

The given series = $\displaystyle \sum_{n=1}^\infty \dfrac{2^n}{n}$

Here, $a_n=\dfrac{2^n}{n}$.

Therefore,

ℓ = lim_n→∞ $\dfrac{a_{n+1}}{a_n}$

= lim_n→∞ $\left( \dfrac{2^{n+1}}{n+1} \times \dfrac{2^{n}}{n} \right)$

= 2 lim_n→∞ $\dfrac{n}{n+1}$

= 2 lim_n→∞ $\dfrac{\frac{n}{n}}{\frac{n+1}{n}}$

= 2 lim_n→∞ $\dfrac{1}{1+\frac{1}{n}}$

= 2 × $\dfrac{1}{1+0}$

= 2.

$\boxed{\color{blue}\textbf{Q}2:}$ Test the convergence of $1+\dfrac{3}{2!}+\dfrac{5}{3!}+\dfrac{7}{4!}+\cdots$.

$\boxed{\color{red}\textbf{Answer:}}$

The given series = $\displaystyle \sum_{n=1}^\infty \dfrac{2n-1}{n!}$

Here, $a_n=\dfrac{2n-1}{n!}$.

Therefore, lim_n→∞ $\dfrac{a_{n+1}}{a_n}$

= lim_n→∞ $\left( \dfrac{2(n+1)-1}{(n+1)!} \times \dfrac{2n-1}{n!} \right)$

= lim_n→∞ $\left( \dfrac{2n+1}{(n+1)(2n-1)} \right)$

= 0.

As this limit is less than 1, by ratio test, the given series converges.

$\boxed{\color{blue}\textbf{Q}3:}$ Test the convergence of $\dfrac{1}{1!}+\dfrac{2^2}{2!}+\dfrac{3^3}{3!}+\cdots$.

$\boxed{\color{red}\textbf{Answer:}}$

The given series = $\displaystyle \sum_{n=1}^\infty \dfrac{n^n}{n!}$.

So a_n = $\dfrac{n^n}{n!}$.

Thus, ℓ = lim_n→∞ $\dfrac{a_{n+1}}{a_n}$

= lim_n→∞ $\left( \dfrac{(n+1)^{n+1}}{(n+1)!} \times \dfrac{n!}{n^n} \right)$

= lim_n→∞ $\left( \dfrac{n+1}{n} \right)^n$

= lim_n→∞ $\left( 1+\dfrac{1}{n} \right)^n$

= e.

Note that the value of e lies between 2 and 3. So the limit ℓ is greater than 1. Hence, by ratio test the given series diverges.

Related Articles:

• Comparison Test for Series: Statement, Examples [Limit Form]
• Convergence of a Series: Definition, Formula, Examples

$\boxed{\color{blue}\textbf{Q}4:}$ Find the values of x, for which the series $x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\cdots$ is convergent or divergent.

$\boxed{\color{red}\textbf{Answer:}}$

The given series = $\displaystyle \sum_{n=1}^\infty \dfrac{x^n}{n}$

Thus a_n = $\dfrac{x^n}{n}$.

Now, ℓ = lim_n→∞ $\dfrac{a_{n+1}}{a_n}$

= lim_n→∞ $\left( \dfrac{x^{n+1}}{n+1} \times \dfrac{n}{x^n} \right)$

= $x \times$ lim_n→∞ $\dfrac{n}{n+1}$

= $x \times 1 = x$.

So by ratio test, the series converges if x<1 and diverges if x>1.

The case x=1. For x=1, the given series is equal to $1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots$ $=\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}$. It is a p-series with p=1. Thus, the series diverges.

Final conclusion: The series ∑xⁿ/n! converges if x<1 and diverges if x≥1.

Homework:

Test the convergence of the following series:

1. $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n 2^n}$
2. $1+\dfrac{1}{1!}+\dfrac{2^2}{2!}+\dfrac{3^3}{3!}+\cdots$ [Ignore the first term].

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