Numerical series.

Series with non-negative terms.

A series $\sum_{n = 1}^{\infty} a_{n}$ with non-negative terms $(a_{n} \geq 0, n \in N)$ converges if and only if the sequence of its partial sums is bounded above, that is, there exists a number $M > 0$ such that for each $n \in N$ the inequality $\sum_{n = 1}^{\infty} a_{n} \leq M$ holds.

Comparison tests.

1. If there exists a number $n_{0}$ such that for all $n \geq n_{0}$ the inequalities $0 \leq a_{n} \leq b_{n}$ hold, then convergence of the series $\sum_{n = 1}^{\infty} b_{n}$ implies convergence of the series $\sum_{n = 1}^{\infty} a_{n}$ , and divergence of the series $\sum_{n = 1}^{\infty} a_{n}$ implies divergence of the series $\sum_{n = 1}^{\infty} b_{n}$ .

2. If $a_{n} \geq 0,,, b_{n} > 0$ for all $n \geq n_{0}$ and there exists a finite non-zero limit $lim_{n \to \infty} \frac{a_{n}}{b_{n}},$ then the series $\sum_{n = 1}^{\infty} a_{n}$ and $\sum_{n = 1}^{\infty} b_{n}$ either both converge or both diverge.

Example 1.

Using the comparison test, investigate the convergence of the series $\sum_{n = 1}^{\infty} \frac{5 + 3 (- 1)^{n + 1}}{2^{n}} .$

Solution.

For odd $n$ , $a_{n} = \frac{8}{2^{n}}$ , and for even $n$ , $a_{n} = \frac{2}{2^{n}}$ . Thus, $0 < \frac{1}{2^{n - 1}} < a_{n} < \frac{1}{2^{n - 3}} .$

Notice that ${\frac{1}{2^{n - 3}}}$ is a geometric progression with $b_{1} = 4$ and $q = \frac{1}{2}$ . The sum of a geometric progression is calculated using the formula $S = \frac{b_{1}}{1 - q}$ . In our case, $S = \frac{4}{1 / 2} = 8$ . Therefore, the series $\sum_{n = 1}^{\infty} \frac{1}{2^{n - 3}}$ converges. By the first convergence test, it follows that $\sum_{n = 1}^{\infty} \frac{5 + 3 (- 1)^{n + 1}}{2^{n}}$ also converges.

Example 2.

Using the comparison test, investigate the convergence of $\sum_{n = 1}^{\infty} \sin \frac{3 + (- 1)^{n}}{n^{2}} .$

Solution.

For odd $n$ , $a_{n} = \sin \frac{2}{n^{2}}$ , and for even $n$ , $a_{n} = \sin \frac{4}{n^{2}}$ . Thus, $0 < \sin \frac{2}{n^{2}} < \sin \frac{3 + (- 1)^{n}}{n^{2}} < \sin \frac{4}{n^{2}} .$

Notice that $lim_{n \to} \frac{\sin \frac{4}{n^{2}}}{\frac{4}{n^{2}}} = 1.$ This means that the series $\sum_{n = 1}^{\infty} \sin \frac{4}{n^{2}}$ and $\sum_{n = 1}^{\infty} \frac{4}{n^{2}}$ converge or diverge simultaneously. Let's consider the series $\frac{4}{n^{2}} :$

$\frac{4}{n^{2}} \leq \frac{4}{n (n - 1)} = - \frac{4}{n} + \frac{4}{n - 1} = 4 (- \frac{1}{n} + \frac{1}{n - 1}) .$

Thus, $\sum_{n = 1}^{\infty} \frac{4}{n^{2}} = 4 (- \frac{1}{2} + 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - . . .)$ $S_{n} = 4 (1 - \frac{1}{n}), S = lim_{n \to \infty} S_{n} = 4 < \infty$ . Therefore, the series converges. Hence, by the convergence test, it follows that $\sum_{n = 1}^{\infty} \sin \frac{3 + (- 1)^{n}}{n^{2}}$ also converges.

Integral convergence test for series.

If the function $f (x)$ is non-negative and decreasing on the interval $[a, + \infty),$ where $a \geq 1,$ then the series $\sum_{n = 1}^{\infty} f (n)$ and the integral $\int_{a}^{+ \infty} f (x) d x$ converge or diverge simultaneously.

Example.

Investigate the convergence of the series $\sum_{n = 1}^{\infty} \frac{1}{(1 + n) \sqrt{n}} .$

Solution.

We will use the integral convergence test. Consider the integral $\int_{1}^{+ \infty} \frac{1}{(1 + x) \sqrt{x}} :$

$\int_{1}^{+ \infty} \frac{1}{(1 + x) \sqrt{x}} = [t = \sqrt{x};,,, d x = 2 t d t] = lim_{A \to + \infty} \int_{1}^{A} \frac{2 t d t}{t (1 + t^{2})} = lim_{A \to + \infty} 2 \arctan t |_{1}^{A} =$ $= 2 \cdot \frac{π}{2} - 2 \cdot \frac{π}{4} = \frac{π}{2} .$

Hence, the integral $\int_{1}^{+ \infty} \frac{1}{(1 + x) \sqrt{x}}$ converges. By the integral convergence test for series, it follows that the series $\sum_{n = 1}^{\infty} \frac{1}{(1 + n) \sqrt{n}}$ also converges.

Cauchy and D'Alembert criteria.

D'Alembert criterion.

If for the series $\sum_{n = 1}^{\infty} a_{n},$ $a_{n} > 0$ $(n \in N),$ there exists a limit $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = λ$ , then if $λ < 1$ the series $\sum_{n = 1}^{\infty} a_{n}$ converges, if $λ > 1$ it diverges. When $λ = 1$ , the series may converge or diverge.

Cauchy criterion.

If for the series $\sum_{n = 1}^{\infty} a_{n},$ $a_{n} \geq 0$ $(n \in N),$ there exists a limit $lim_{n \to \infty} \sqrt[n]{a_{n}} = λ$ , then if $λ < 1$ the series converges, if $λ > 1$ it diverges.

Examples.

1. Investigate the convergence of the series $\sum_{n = 1}^{\infty} a_{n}$ using the D'Alembert criterion.

$a_{n} = \frac{n^{10}}{(n + 1)!}$

Solution.

$a_{n + 1} = \frac{(n + 1)^{10}}{(n + 1)! (n + 2)}$

$lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = lim_{n \to \infty} \frac{(n + 1)^{10} (n + 1)!}{(n + 1)! (n + 2) n^{10}} = lim_{n \to \infty} \frac{(n + 1)^{10}}{n^{10} (n + 2)} =$ $lim_{n \to \infty} \frac{n^{10} (1 + 1 / n)^{10}}{n^{10} (n + 2)} = lim_{n \to \infty} \frac{(1 + 1 / n)^{10}}{n + 2} = 0 < 1.$

Therefore, the series converges.

2. Investigate the convergence of the series $\sum_{n = 1}^{\infty} a_{n}$ using the Cauchy criterion.

$a_{n} = 3^{n + 1} {(\frac{n + 2}{n + 3})}^{n^{2}}$

Solution.

$lim_{n \to \infty} \sqrt[n]{a_{n}} = lim_{n \to \infty} 3 \sqrt[n]{3} {(\frac{n + 2}{n + 3})}^{n} = lim_{n \to \infty} 3 \sqrt[n]{3} {(\frac{n + 3 - 1}{n + 3})}^{n} =$ $lim_{n \to \infty} 3 \sqrt[n]{3} {({(1 - \frac{1}{n + 3})}^{- (n + 3) + 3})}^{- 1} = \frac{3}{e} lim_{n \to \infty} {(1 + \frac{1}{n + 3})}^{- 3} = \frac{3}{e} .$

Raabe and Gauss criteria.

Raabe criterion.

If $a_{n} > 0,, (n \in N)$ and there exists a limit $lim_{n \to \infty} n (\frac{a_{n}}{a_{n + 1}} - 1) = q$ , then if $q > 1$ the series $\sum_{n = 1}^{\infty} a_{n}$ converges and if $q < 1$ it diverges.

Gauss criterion.

If $a_{n} > 0,,, (n \in N)$ and $\frac{a_{n}}{a_{n + 1}} = α + \frac{β}{n} + \frac{γ_{n}}{n^{1 + δ}},$ where $| γ_{n} | < C < δ > 0,$ then:

a) if $α > 1$ the series $\sum_{n = 1}^{\infty} a_{n}$ converges, and if $α < 1$ it diverges.

b) if $α = 1$ the series converges if $β > 1$ and diverges if $β \leq 1$ .

Examples.

1. Investigate the convergence of the series $\sum_{n = 1}^{\infty} a_{n}$ using the Raabe or Gauss criterion.

$a_{n} = \frac{(2 n - 1)!!}{(2 n)!!}$

Solution.

We will use the Raabe criterion.

$a_{n + 1} = \frac{(2 n + 1)!!}{(2 n + 2)!!} = \frac{(2 n - 1)!! (2 n + 1)}{(2 n)!! (2 n + 2)} = a_{n} \frac{2 n + 1}{2 n + 2} .$

$lim_{n \to \infty} n (n (\frac{a_{n}}{a_{n + 1} - 1})) = lim_{n \to \infty} n (\frac{2 n + 2}{2 n + 1} - 1) = lim_{n \to \infty} n (\frac{2 n + 2 - 2 n - 1}{2 n + 1}) =$ $lim_{n \to \infty} \frac{n}{2 n + 1} = \frac{1}{2} < 1.$

Therefore, the series diverges.

2. Investigate the convergence of the series $\sum_{n = 1}^{\infty} a_{n}$ using the Raabe or Gauss criterion.

$a_{n} = {(\frac{(a + 1) (a + 2) . . . (a + n)}{(b + 1) (b + 2) . . . (b + n)})}^{α}$

Solution.

We will apply the Gauss criterion.

$a_{n + 1} = {(\frac{(a + 1) (a + 2) . . . (a + n + 1)}{(b + 1) (b + 2) . . . (b + n + 1)})}^{α} = a_{n} {(\frac{(a + n + 1}{b + n + 1})}^{α} .$

$\frac{a_{n}}{a_{n + 1}} = {(\frac{b + n + 1}{a + n + 1})}^{α} = {(1 + \frac{b - a}{a + n + 1})}^{α} = 1 + α \frac{b - a}{a + n + 1} +$ $α (α - 1) \frac{1}{2} {(\frac{b - a}{a + n + 1})}^{2} + . . .$

Therefore, the series converges if $α (b - 1) > 1$ and diverges if $α (b - a) \leq 1.$

Alternating series.

A series $\sum_{n = 1}^{\infty} (- 1)^{n - 1} a_{n} = a_{1} - a_{2} + a_{3} + . . .,$ where $a_{n} \geq 0$ or $a_{n} \leq 0$ for all $n \in N$ is called an alternating series.

Leibniz criterion.

If $lim_{n \to \infty} a_{n} = 0$ and for each $n \in N$ it holds that $a_{n} \geq a_{n + 1} > 0$ , then the series $\sum_{n = 1}^{\infty} (- 1)^{n - 1} a_{n}$ converges. Moreover, $| S - S_{n} | \leq a_{n + 1},$ where $S$ and $S_{n}$ are the sum and $n$ -th partial sum of the series $\sum_{n = 1}^{\infty} (- 1)^{n - 1} a_{n} .$

Examples.

1. Investigate the convergence of the series $\sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{\sqrt[3]{n + 1}} .$

Solution.

Let's use the Leibniz criterion. Since the sequence ${\frac{1}{\sqrt[3]{n + 1}}}$ decreases and tends to 0 as $n \to \infty,$ by the Leibniz criterion, the series converges.