Calculus II Home

## Series with Positive Terms

In this section we study four tests of convergence for series with positive terms: The Integral Test, the $$p$$-series Test, the Comparison Test, the Limit Comparison Test.

The Integral Test: Let $$a_n=f(n)$$ for all integers $$n\geq 1$$ where $$f$$ is a continuous, positive, and decreasing function on $$[1,\infty)$$. Then $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is convergent if and only if $$\displaystyle\int_1^{\infty} f(x)\, dx$$ is convergent.

Since $$f$$ is decreasing, $0< a_2+a_3+\cdots+a_n\leq \int_1^n f(x)\, dx \implies a_1< s_n\leq a_1+\int_1^n f(x)\, dx.$ Taking limit as $$n\to \infty$$, $a_1\leq \lim_{n\to \infty} s_n\leq a_1+ \lim_{n\to \infty}\int_1^n f(x)\, dx \implies a_1\leq \sum_{n=1}^{\infty} a_n \leq a_1+ \int_1^{\infty} f(x)\, dx.$ If $$\displaystyle\int_1^{\infty} f(x)\, dx$$ is convergent, then $a_1\leq \sum_{n=1}^{\infty} a_n \leq a_1+ \int_1^{\infty} f(x)\, dx< \infty.$ Similarly \begin{align*} &\int_1^n f(x)\, dx \leq a_1+a_2+\cdots+a_{n-1}=s_{n-1}\\ \implies & \int_1^{\infty} f(x)\, dx=\lim_{n\to \infty}\int_1^n f(x)\, dx \leq \lim_{n\to \infty} s_{n-1}=\sum_{n=1}^{\infty} a_n. \end{align*} If $$\displaystyle\int_1^{\infty} f(x)\, dx$$ is divergent, then $\infty=\int_1^{\infty} f(x)\, dx \leq \sum_{n=1}^{\infty} a_n \implies \sum_{n=1}^{\infty} a_n=\infty.$

Example. Use the Integral Test to show that $$\displaystyle \sum_{n=1}^{\infty} \frac{2n}{n^4+3}$$ is convergent.

Solution. First note that $$f(x)=\displaystyle\frac{2x}{x^4+3}$$ is a continuous, positive, and decreasing function on $$[1,\infty)$$. \begin{align*} \int_1^{\infty} f(x)\, dx &=\int_1^{\infty} \frac{2x}{x^4+3}\, dx\\ &=\lim_{t\to \infty}\int_1^t \frac{2x}{x^4+3}\, dx\\ &=\lim_{t\to \infty} \left.\frac{1}{\sqrt{3}}\tan^{-1} \left( \frac{x^2}{\sqrt{3}} \right) \right\vert_1^t \;\;\left(\text{Hint. } u=x^2 \right)\\ &=\lim_{t\to \infty} \frac{1}{\sqrt{3}} \left[ \tan^{-1} \left( \frac{t^2}{\sqrt{3}} \right) - \tan^{-1}\left( \frac{1}{\sqrt{3}} \right) \right]\\ &= \frac{1}{\sqrt{3}} \left[ \frac{\pi}{2}-\frac{\pi}{6} \right]\\ &=\frac{\pi}{3\sqrt{3}}. \end{align*} Since $$\displaystyle\int_1^{\infty} \frac{2x}{x^4+3}\, dx$$ is convergent, $$\displaystyle \sum_{n=1}^{\infty} \frac{2n}{n^4+3}$$ is convergent by the Integral Test.

The $$p$$-series Test: The $$p$$-series $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$$ is convergent if $$p>1$$ and divergent if $$p\leq 1$$.

Let $$p > 1$$. Then $$f(x)=\frac{1}{x^p}$$ is a continuous, positive, and decreasing function on $$[1,\infty)$$. Also $$\displaystyle \int_1^{\infty} \frac{1}{x^p}\, dx$$ is convergent by the integral $$p$$-test. Then by the Integral Test, $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$$ is convergent.
By similar arguments we can prove that $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$$ is divergent when $$0 < p\leq 1$$. Finally for $$p\leq 0$$, $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$$ is divergent by the Divergence Test.

Example.

1. $$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$$ is convergent by the $$p$$-series Test with $$p=2 > 1$$.
2. $$\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$$ is divergent by the $$p$$-series Test with $$p=\frac{1}{2}\leq 1$$.

The Comparison Test: Suppose $$\displaystyle\sum_{n=1}^{\infty} a_n$$ and $$\displaystyle\sum_{n=1}^{\infty} b_n$$ are two series with positive terms.

1. If $$a_n\leq b_n$$ for all integers $$n\geq 1$$ and $$\displaystyle\sum_{n=1}^{\infty} b_n$$ is convergent, then $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is also convergent.
2. If $$a_n\leq b_n$$ for all integers $$n\geq 1$$ and $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is divergent, then $$\displaystyle\sum_{n=1}^{\infty} b_n$$ is also divergent.

Note that we often compare a series with the $$p$$-series $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$$ or the geometric series $$\displaystyle\sum_{n=1}^{\infty} ar^{n-1}$$.

Example.

1. Use the Comparison Test to show that $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{2n^5+1}}$$ is convergent.

Solution. For all integers $$n\geq 1$$, $0 < \frac{1}{\sqrt[3]{2n^5+1}}\leq \frac{1}{\sqrt[3]{2n^5}}=\frac{1}{\sqrt[3]{2}n^{\frac{5}{3}}}.$ $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{\frac{5}{3}}}$$ is convergent by the $$p$$-series Test with $$p=\frac{5}{3} > 1$$. Then $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{2}n^{\frac{5}{3}}} =\frac{1}{\sqrt[3]{2}} \sum_{n=1}^{\infty} \frac{1}{n^{\frac{5}{3}}}$$ is convergent. Finally by the Comparison Test, $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{2n^5+1}}$$ is convergent.

2. Use the Comparison Test to show that $$\displaystyle \sum_{n=1}^{\infty} \frac{3^n}{2^n-1}$$ is divergent.

Solution. For all integers $$n\geq 1$$, $\frac{3^n}{2^n-1}\geq \frac{3^n}{2^n}=\left( \frac{3}{2}\right)^n > 0.$ $$\displaystyle \sum_{n=1}^{\infty} \left( \frac{3}{2}\right)^n$$ is divergent as a geometric series with the common ratio $$r=\frac{3}{2} \geq 1$$. Then by the Comparison Test, $$\displaystyle \sum_{n=1}^{\infty} \frac{3^n}{2^n-1}$$ is divergent.

The Limit Comparison Test: Suppose $$\displaystyle\sum_{n=1}^{\infty} a_n$$ and $$\displaystyle\sum_{n=1}^{\infty} b_n$$ are two series with positive terms. If $$\displaystyle\lim_{n\to \infty} \frac{a_n}{b_n}$$ is a positive number, then either both series converge or both series diverge.

Example. Use the Limit Comparison Test to show that $$\displaystyle \sum_{n=1}^{\infty} \frac{\sqrt{n^4-1}}{2n^3+n^2}$$ is divergent.

Solution. First we construct a series with positive terms by keeping the highest degree terms of the numerator and denominator of the given series: $\displaystyle \sum_{n=1}^{\infty} \frac{\sqrt{n^4}}{n^3}= \displaystyle \sum_{n=1}^{\infty}\frac{n^2}{n^3}= \displaystyle \sum_{n=1}^{\infty}\frac{1}{n}.$ \begin{align*} \lim_{n\to \infty} \displaystyle\frac{\frac{\sqrt{n^4-1}}{2n^3+n^2}}{\frac{1}{n}} &= \lim_{n\to \infty} \frac{\sqrt{n^4-1}}{2n^3+n^2}\cdot\frac{n}{1}\\ &= \lim_{n\to \infty} \frac{n\sqrt{n^4-1}/n^3}{(2n^3+n^2)/n^3}\\ &= \lim_{n\to \infty} \frac{\sqrt{1-\frac{1}{n^4}}}{2+\frac{1}{n}}\\ &= \frac{1}{2}>0. \end{align*} Since the harmonic series $$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$$ is divergent, $$\displaystyle \sum_{n=1}^{\infty} \frac{\sqrt{n^4-1}}{2n^3+n^2}$$ is divergent by the Limit Comparison Test.

Last edited