Comparison Test for Series: Statement, Examples [Limit Form]

The comparison test for series is used to check whether the series is convergent or not. Comparison test is useful when one can compare the given series with another series whose convergence is known. In this article, we learn about comparison test of a series with its limit form and some solved examples.

Comparison Test for Series

Let $\displaystyle \sum_{n=1}^\infty a_n$ and $\displaystyle \sum_{n=1}^\infty b_n$ be two series of non-negative real numbers. Suppose that the following holds.

∑a_n ≤ ∑b_n.

Then we have:

1. If ∑b_n converges, then ∑a_n converges.
2. If ∑a_n diverges, then ∑b_n diverges.

Related Articles: Convergence of a series

Solved Problems

Before we solve few problems using comparison test, let us recall the geometric series and p-series.

• A geometric series a+ar+ar²+ar³+… converges if -1<r<1; otherwise diverges.
• A p-series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^p}$ converges if p>1; otherwise diverges.

$\boxed{\color{blue}\textbf{Problem 1:}}$ Test the convergence of $\displaystyle \sum_{n=1}^\infty \dfrac{\sin nx}{n^2}$ (x>0).

$\boxed{\color{red}\textbf{Solution:}}$ For any real number x, we know that

sin(nx) ≤ 1 for all n.

⇒ $\dfrac{\sin nx}{n^2} \leq \dfrac{1}{n^2}$ for all n.

Thus, we get that

$\displaystyle \sum_{n=1}^\infty \dfrac{\sin nx}{n^2} \leq \displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}$

As $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}$ converges (because it is a p-series with p=2>1), by comparison test we conclude that the series $\displaystyle \sum_{n=1}^\infty \dfrac{\sin nx}{n^2}$ converges.

Limit Form of Comparison Test

Let ∑a_n and ∑b_n be two series of positive real numbers. Let

ℓ = lim_n→∞ $\dfrac{a_n}{b_n}$.

If ℓ is a finite positive number, that is, 0<ℓ<∞, then the two series ∑a_n and ∑b_n behave similarly. In other words, if ∑a_n converges then ∑b_n converges and vice-versa (also, if ∑a_n diverges then ∑b_n diverges and vice-versa).

Question-Answer

$\boxed{\color{blue}\textbf{Q}1:}$ Test the convergence of $\displaystyle \sum_{n=1}^\infty \dfrac{1}{2n+5}$.

$\boxed{\color{red}\textbf{Answer:}}$ Here, $a_n=\dfrac{1}{2n+5}$.

Let us consider the series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}$. Thus, $b_n=\dfrac{1}{n}$.

Now, ℓ = lim_n→∞ $\dfrac{a_n}{b_n}$

= lim_n→∞ $\dfrac{n}{2n+5}$

= lim_n→∞ $\dfrac{\frac{n}{n}}{\frac{2n+5}{n}}$

= lim_n→∞ $\dfrac{1}{2+\frac{5}{n}}$

= 1/2.

As the value of ℓ is a 1/2, that is, 0<ℓ<∞, then the two series ∑a_n and ∑b_n behave similarly. From p-series, we know ∑ 1/n diverges. Therefore, by limit form of comparison test, the given series $\sum_{n=1}^\infty \frac{1}{2n+5}$ also diverges.

$\boxed{\color{blue}\textbf{Q}2:}$ Test the convergence of $\displaystyle \sum_{n=1}^\infty \dfrac{1}{2n^2+5n}$.

$\boxed{\color{red}\textbf{Answer:}}$ Here, $a_n=\dfrac{1}{2n^2+5n}$.

We compare it with the series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}$, so $b_n=\dfrac{1}{n^2}$.

Now, ℓ = lim_n→∞ $\dfrac{a_n}{b_n}$

= lim_n→∞ $\dfrac{n^2}{2n^2+5n}$

= lim_n→∞ $\dfrac{\frac{n^2}{n^2}}{\frac{2n^2+5n}{n^2}}$

= lim_n→∞ $\dfrac{1}{2+\frac{5}{n}}$

= 1/2.

Since 0< ℓ=1/2 <∞, the two series ∑a_n and ∑b_n converge or diverge together. From p-series, we know ∑ 1/n² converges. Therefore, by limit form of comparison test, the series $\sum_{n=1}^\infty \frac{1}{2n^2+5n}$ converges.

$\boxed{\color{blue}\textbf{Q}3:}$ Test the convergence of $\displaystyle \sum_{n=1}^\infty \sin \left(\dfrac{1}{n} \right)$.

$\boxed{\color{red}\textbf{Answer:}}$ Here, $a_n=\sin \left(\dfrac{1}{n} \right)$.

We compare it with the series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n}$, so $b_n=\dfrac{1}{n}$.

Now, ℓ = lim_n→∞ $\dfrac{a_n}{b_n}$

= lim_n→∞ $\dfrac{\sin \left(\frac{1}{n} \right)}{\frac{1}{n}}$

= lim_t→0 $\dfrac{\sin t}{t}$ where t=1/n. so t→0 when n→∞.

= 1.

Since 0< ℓ=1 <∞, we have ∑a_n and ∑b_n converge or diverge together. From p-series, we know ∑ 1/n diverges. Therefore, by limit form of comparison test, $\sum_{n=1}^\infty \sin \left(\frac{1}{n} \right)$ diverges.

Homework:

Test the convergence of the following series:

• $\sum_{n=1}^\infty \dfrac{1}{2n^2+5}$
• $\sum_{n=1}^\infty \dfrac{n}{n^4+1}$
• $\sum_{n=1}^\infty \dfrac{1}{n} \sin \left(\dfrac{1}{n} \right)$

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