In the infinite series notes you saw the series expansions of the sine and cosine functions. Here they are again along with the summation notation. Remember that there's nothing especially complicated about Σ notation; it's just a way of capturing what's going on in the series without having to write all or several of its terms.

$$ \begin{align} sin(x) &= \frac{x^1}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \dots = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{(2n + 1)!} \\ \\ cos(x) &= \frac{x^0}{0!} - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \dots = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \end{align}$$

These are called **alternating series** because of the alternation in the sign of each term.

To describe such an alternation in summation notation we employ the properties of powers of negative numbers:

- $(-1)^1 = -1$
- $(-1)^2 = 1$
- $(-1)^3 = -1,$ and so on.

It's also possible to produce an alternation of sign using the trigonometric functions **sin(x)** and **cos(x)**, like this:

- $cos(n\pi) = 1$ for even n and -1 for odd n
- $sin\left(\frac{n\pi}{2}\right) = 1$ for odd n and -1 for even n

Here is an example of an alternating series, the so-called **alternating harmonic series**. This is just the harmonic series with alternating signs of the terms. The terms alternate on either side of zero as they decrease to zero (blue graph).

The accumulating sum (red graph) converges to a limit of approximately 0.69, but oscillates about that line. It's non-trivial to determine the actual *sum* of this series, but it does converge.

The alternating series test, proved below the next box, is very simple.

It says that if, as n→∞, the terms of an alternating series decrease to zero, then the series converges.

Remember, that is NOT necessarily true for non-alternating series. For a non-alternating series, it is not enough that the size of the term diminishes; that series still may not converge.

We can think of many non-alternating, divergent series with decreasing terms, like the harmonic series:

$$\sum_{n = 1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \:\dots $$

In a way, the alternating series test (AST) makes life a lot easier. It's enough that the absolute value of the terms decrease to zero for an alternating series to converge – pretty simple.

If the terms of an alternating series, $\sum_{n = 1}^\infty \,(-1)^n a_n$

(1) are **decreasing**, and (2) if the limit of the size of the terms, as n → ∞ is zero:

then the series converges.

First, take a look at the alternating series on a number line. We'll let the terms of the series be $a_1, a_2, a_3, \dots$ and so on, and the partial sums will be $s_1 = a_1, \; s_2 = a_1 + a_2, \; s_3 = a_1 + a_2 + a_3,\; \dots$ and so on.

Let's look first at the **even** sums. The first one (first green arrow in the diagram) is

$$s_2 = a_1 - a_2 \gt 0 \; \leftarrow \; \text{ because } a_1 \gt a_2$$

The second and third are

$$ \begin{align} s_4 &= s_2 + (a_3 - a_4) \ge s_2 \; \leftarrow \; \text{ because } a_3 \gt a_4 \\[4pt] s_6 &= s_4 + (a_5 - a_6) \ge s_4 \; \leftarrow \; \text{ because } a_5 \gt a_6 \end{align}$$

... and the pattern emerges. In the notation below, **2n** is always an even number, **2n-2** is the next smaller even number and **2n-1** is the next smaller odd number:

$$s_{2n} = s_{2n - 2} + (a_{2n - 1} - a_{2n}) \ge s_{2n - 2}$$

Now it's clear that these sums continue to grow, so that

$$0 \le s_2 \le s_4 \le s_6 \le \dots \le s_{2n} \le \dots$$

By regrouping terms with some parenthesis, we can also write **s _{2n}** as

$$s_{2n} = a_1 - (a_2 - a_1) - (a_4 - a_3) - \dots - (a_{2n - 2} - a_{2n - 1}) - a_{2n}$$

Now because our terms are decreasing, every difference in parenthesis is positive. The sequence of terms **{s _{n}}** is positive, increasing and bounded from above by the sum of the series,

$$\lim_{n\to\infty} s_{2n} = s$$

Now because $s_{2n + 1} = s_{2n} + a+{2n + 1},$ we can find the limit of the odd partial sums:

$$\lim_{n\to\infty} s_{2n + 1} = \lim_{n\to\infty} (s_{2n} + a_{2n + 1})$$

What we've shown is that the limits of the even and odd partial sums is the same. This is called "squeezing." If the limits from above and below are **s**, the sum of the series, then the series must converge to **s**.

The difference in convergence between the alternating harmonic series,

$$\sum_{n = 1}^\infty \, (-1)^n \frac{1}{n}$$

which converges, and the harmonic series

$$\sum_{n = 1}^\infty \, \frac{1}{n}$$

which doesn't, hints at a subtle differentiation in how series converge. In fact, the alternating harmonic series is called "**conditionally convergent**," the condition being that its terms alternate sign.

On the other hand, a series like the convergent p-series,

$$\sum_{n = 1}^\infty \, \frac{1}{n^2}$$

and its alternating series counterpart,

$$\sum_{n = 1}^\infty \, (-1)^n \frac{1}{n^2}$$

*both* converge, and the alternating version is said to be "**absolutely convergent**." Here are the rules for absolute and conditional convergence:

Test for **absolute convergence**

Test for **conditional convergence**

**Divergence** – the series diverges by any one of the tests for convergence, or the divergence test.

Does the series $\sum_{n = 0}^\infty \, (-1)^{n + 1} \frac{2n + 3}{3n + 4}$ converge?

**Solution****n** are negative. To test whether it converges, we ask whether the term, without the alternating part (-1)^{n+1}.

$$\lim_{n\to\infty} \, \frac{2n + 3}{3n + 4} = \frac{2}{3} \neq 0$$

The term does not tend toward zero, so this series does not converge. We could also say that it diverges by the divergence test. Either way, it diverges.

Does the series $\sum_{n = 0}^\infty \, \frac{(-2)^n}{3^n}$ converge?

**Solution****(xy) ^{n} = x^{n}·y^{n}**:

$$\sum_{n = 0}^\infty \, (-1)^n \left( \frac{2}{3} \right)^n$$

Now we need to find whether the limit of the term without the alternation is zero:

$$\lim_{n\to\infty} \, \left( \frac{2}{3} \right)^n = 0$$

2/3 < 1, so the limit is zero, and this alternating series **converges** by the AST.

Decide whether the following series converge.

1. | $$\sum_{n = 1}^\infty \, \frac{(-1)^n}{4 + 3n}$$ | |

2. | $$\sum_{n = 1}^\infty \, \frac{(-1)^n}{n^3 + 2n + 2}$$ |

3. | $$\sum_{n = 1}^\infty \, \frac{n \cdot cos(n\pi)}{n^2 + 1}$$ | |

4. | $$\sum_{n = 1}^\infty \, \frac{(-1)^n n^2}{n^2 + 1}$$ |

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