# Comparison Tests

## Solved Problems

### Example 3.

Determine whether the series $\sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{n^3} - 3}}}$ converges or diverges.

Solution.

Use the comparison test. Note that $${n^3} - 3 \lt {n^3}$$ for all integers $$n.$$ Then

$\frac{1}{{{n^3} - 3}} \gt \frac{1}{{{n^3}}},\;\; \Rightarrow \frac{{{n^2}}}{{{n^3} - 3}} \gt \frac{{{n^2}}}{{{n^3}}} = \frac{1}{n}.$

Since $$\sum\limits_{n = 1}^\infty {\frac{1}{n}}$$ is the harmonic series, it diverges. Hence, the given series also diverges by the comparison test.

### Example 4.

Determine whether the series $\sum\limits_{n = 1}^\infty {\frac{{3n - 1}}{{2{n^3} - 4n + 5}}}$ converges or diverges.

Solution.

Use the limit comparison test. We will compare with the convergent $$p$$-series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}.$$ Then

$L = \lim\limits_{n \to \infty } \frac{{\frac{{3n - 1}}{{2{n^3} - 4n + 5}}}}{{\frac{1}{{{n^2}}}}} = \lim\limits_{n \to \infty } \frac{{\left( {3n - 1} \right){n^2}}}{{2{n^3} - 4n + 5}} = \lim\limits_{n \to \infty } \frac{{3{n^3} - {n^2}}}{{2{n^3} - 4n + 5}}.$

Divide the nominator and denominator by $${n^3}:$$

$L = \lim\limits_{n \to \infty } \frac{{\frac{{3\cancel{n^3}}}{{\cancel{n^3}}} - \frac{{{n^2}}}{{{n^3}}}}}{{\frac{{2\cancel{n^3}}}{{\cancel{n^3}}} - \frac{{4n}}{{{n^3}}} + \frac{5}{{{n^3}}}}} = \lim\limits_{n \to \infty } \frac{{3 - \frac{1}{n}}}{{2 - \frac{4}{{{n^2}}} + \frac{5}{{{n^3}}}}} = \frac{3}{2}.$

Hence, the given series converges by the limit comparison test.

### Example 5.

Determine whether $\sum\limits_{n = 1}^\infty {\frac{{\sqrt n }}{{2{n^2} + n + 5}}}$ is convergent.

Solution.

We will compare this series with the convergent series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^{\frac{3}{2}}}}}} .$$ Then we have

$L = \lim\limits_{n \to \infty } \frac{{\frac{{\sqrt n }}{{2{n^2} + n + 5}}}}{{\frac{1}{{{n^{\frac{3}{2}}}}}}} = \lim\limits_{n \to \infty } \frac{{\sqrt n \cdot {n^{\frac{3}{2}}}}}{{2{n^2} + n + 5}} = \lim\limits_{n \to \infty } \frac{{{n^2}}}{{2{n^2} + n + 5}} = \lim\limits_{n \to \infty } \frac{{\frac{{\cancel{n^2}}}{{\cancel{n^2}}}}}{{\frac{{2\cancel{n^2}}}{{\cancel{n^2}}} + \frac{n}{{{n^2}}} + \frac{5}{{{n^2}}}}} = \lim\limits_{n \to \infty } \frac{1}{{2 + \frac{1}{n} + \frac{5}{{{n^2}}}}} = \frac{1}{2}.$

Consequently, the given series is convergent by the limit comparison test.

### Example 6.

Determine whether the series $\sum\limits_{n = 1}^\infty {\frac{n}{{{n^2} - 2n - 3}}}$ converges or diverges.

Solution.

We use the limit comparison test. Compare $$\sum\limits_{n = 1}^\infty {\frac{n}{{{n^2} - 2n - 3}}}$$ with the divergent harmonic series $$\sum\limits_{n = 1}^\infty {\frac{1}{n}}.$$ To calculate the limit, we divide the nominator and denominator by $${n^2}.$$ Then we have

$L = \lim\limits_{n \to \infty } \frac{{\frac{n}{{{n^2} - 2n - 3}}}}{{\frac{1}{n}}} = \lim\limits_{n \to \infty } \frac{{{n^2}}}{{{n^2} - 2n - 3}} = \lim\limits_{n \to \infty } \frac{{\frac{{\cancel{n^2}}}{{\cancel{n^2}}}}}{{\frac{{\cancel{n^2}}}{{\cancel{n^2}}} - \frac{{2n}}{{{n^2}}} - \frac{3}{{{n^2}}}}} = \lim\limits_{n \to \infty } \frac{1}{{1 - \frac{2}{n} - \frac{3}{{{n^2}}}}} = 1.$

Thus, the given series diverges by the limit comparison test.

### Example 7.

Determine whether the series

$\frac{1}{{\sqrt 2 }} + \frac{1}{{2\sqrt 3 }} + \frac{1}{{3\sqrt 4 }} + \ldots + \frac{1}{{n\sqrt {n + 1} }} + \ldots$

converges or diverges.

Solution.

We use the limit comparison test with the convergent $$p$$-series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^{\frac{3}{2}}}}}} .$$ Find the value of the limit:

$L = \lim\limits_{n \to \infty } \frac{{\frac{1}{{n\sqrt {n + 1} }}}}{{\frac{1}{{{n^{\frac{3}{2}}}}}}} = \lim\limits_{n \to \infty } \frac{{\cancel{n}\sqrt n }}{{\cancel{n}\sqrt {n + 1} }} = \lim\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt {n + 1} }} = \lim\limits_{n \to \infty } \sqrt {\frac{n}{{n + 1}}} = \lim\limits_{n \to \infty } \sqrt {\frac{{\frac{\cancel{n}}{\cancel{n}}}}{{\frac{\cancel{n}}{\cancel{n}} + \frac{1}{n}}}} = \lim\limits_{n \to \infty } \sqrt {\frac{1}{{1 + \frac{1}{n}}}} = 1.$

We see that the given series converges by the limit comparison test.