Comparison Tests

The Nth term test, generally speaking, does not guarantee convergence of a series. Convergence or divergence of a series is proved using sufficient conditions. The comparison tests we consider below are just the sufficient conditions of convergence or divergence of series.

The Comparison Tests

Let \(\sum\limits_{n = 1}^\infty {{a_n}} \) and \(\sum\limits_{n = 1}^\infty {{b_n}} \) be series such that 0 < a_n ≤ b_n for all n. Then the following comparison tests hold:

If \(\sum\limits_{n = 1}^\infty {{b_n}} \) is convergent, then \(\sum\limits_{n = 1}^\infty {{a_n}} \) is also convergent;
If \(\sum\limits_{n = 1}^\infty {{a_n}} \) is divergent, then \(\sum\limits_{n = 1}^\infty {{b_n}} \) is also divergent.

The Limit Comparison Tests

Let \(\sum\limits_{n = 1}^\infty {{a_n}} \) and \(\sum\limits_{n = 1}^\infty {{b_n}}\) be series such that \({a_n}\) and \({b_n}\) are positive for all \(n.\) Then the following limit comparison tests are valid:

If \(0 \lt \lim\limits_{n \to \infty } {\frac{{{a_n}}}{{{b_n}}}} \lt \infty ,\) then \(\sum\limits_{n = 1}^\infty {{a_n}} \) and \(\sum\limits_{n = 1}^\infty {{b_n}}\) are both convergent or both divergent;
If \(\lim\limits_{n \to \infty } {\frac{{{a_n}}}{{{b_n}}}} = 0,\) then \(\sum\limits_{n = 1}^\infty {{b_n}}\) convergent implies that the series \(\sum\limits_{n = 1}^\infty {{a_n}}\) is also convergent;
If \(\lim\limits_{n \to \infty } {\frac{{{a_n}}}{{{b_n}}}} = \infty,\) then \(\sum\limits_{n = 1}^\infty {{b_n}}\) divergent implies that the series \(\sum\limits_{n = 1}^\infty {{a_n}}\) is also divergent.

The so-called \(p\)-series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \) converges for \(p \gt 1\) and diverges for \(0 \lt p \le 1.\)

Solved Problems

Example 1.

Determine whether \[\sum\limits_{n = 1}^\infty {\frac{{{e^{\frac{1}{n}}}}}{{{n^2}}}}\] converges or diverges.

Solution.

We easily can see that \({e^{\frac{1}{n}}} \le e\) for \(n \gt 1.\) Then, by the comparison test,

\[\sum\limits_{n = 1}^\infty {\frac{{{e^{\frac{1}{n}}}}}{{{n^2}}}} \le \sum\limits_{n = 1}^\infty {\frac{e}{{{n^2}}}} = e\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} .\]

Since the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}\) is convergent as a \(p\)-series with the power \(p = 2,\) the original series also converges.

Example 2.

Determine whether the series \[\sum\limits_{n = 1}^\infty {\frac{{{n^2} - 1}}{{{n^4}}}}\] converges or diverges.

Solution.

We use the comparison test. Note that

\[\frac{{{n^2} - 1}}{{{n^4}}} \lt \frac{{{n^2}}}{{{n^4}}} = \frac{1}{{{n^2}}}\]

for all positive integers \(n.\) As \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) is a \(p\)-series with \(p = 2 \gt 1\), it converges. Hence, the given series also converges by the comparison test.

See more problems on Page 2.