Calculus

Infinite Sequences and Series

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Comparison Tests

The \(N\)th 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 \(0 \lt {a_n} \le {b_n}\) for all \(n.\) Then the following comparison tests hold:

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:

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

Click or tap a problem to see the solution.

Example 1

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

Example 2

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

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.

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