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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.


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