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:

- 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 \(0 \lt \lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} \lt \infty ,\) then \(\sum\limits_{n = 1}^\infty {{a_n}} \) and \(\sum\limits_{n = 1}^\infty {{b_n}}\) are both convergent or both divergent;
- If \(\lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} = 0,\) then \(\sum\limits_{n = 1}^\infty {{b_n}}\) convergent implies that the series \(\sum\limits_{n = 1}^\infty {{a_n}}\) is also convergent;
- If \(\lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} = \infty,\) 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 \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^p}}}\normalsize} \) 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 \(\sum\limits_{n = 1}^\infty {\large\frac{{{e^{\frac{1}{n}}}}}{{{n^2}}}\normalsize}\) converges or diverges.### Example 2

Determine whether the series \(\sum\limits_{n = 1}^\infty {\large\frac{{{n^2} – 1}}{{{n^4}}}\normalsize}\) converges or diverges.### Example 3

Determine whether the series \(\sum\limits_{n = 1}^\infty {\large\frac{{{n^2}}}{{{n^3} – 3}}\normalsize}\) converges or diverges.### Example 4

Determine whether the series \(\sum\limits_{n = 1}^\infty {\large\frac{{3n – 1}}{{2{n^3} – 4n + 5}}\normalsize}\) converges or diverges.### Example 5

Determine whether \(\sum\limits_{n = 1}^\infty {\large\frac{{\sqrt n }}{{2{n^2} + n + 5}}\normalsize} \) is convergent.### Example 6

Determine whether the series \(\sum\limits_{n = 1}^\infty {\large\frac{n}{{{n^2} – 2n – 3}}\normalsize}\) converges or diverges.### Example 7

Determine whether the series### Example 1.

Determine whether \(\sum\limits_{n = 1}^\infty {\large\frac{{{e^{\frac{1}{n}}}}}{{{n^2}}}\normalsize}\) converges or diverges.Solution.

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

\[

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

### Example 2.

Determine whether the series \(\sum\limits_{n = 1}^\infty {\large\frac{{{n^2} – 1}}{{{n^4}}}\normalsize}\) converges or diverges.Solution.

We use the comparison test. Note that \({\large\frac{{{n^2} – 1}}{{{n^4}}}\normalsize} \) \( {\large\frac{{{n^2}}}{{{n^4}}}\normalsize} \) \(= {\large\frac{1}{{{n^2}}}\normalsize}\) for all positive integers \(n.\) As \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}}\normalsize} \) is a \(p\)-series with \(p = 2 \gt 1\), it converges. Hence, the given series also converges by the comparison test.