Alternating series

In mathematics, an alternating series is an infinite series of the form

$${\displaystyle \sum _{�=0}^{\mathrm{\infty }}(-1{)}^{�}{�}_{�}}$$

or

$${\displaystyle \sum _{�=0}^{\mathrm{\infty }}(-1{)}^{�+1}{�}_{�}}$$

with a_n > 0 for all n. The signs of the general terms alternate between positive and negative. Like any series, an alternating series converges if and only if the associated sequence of partial sums converges.

Examples[edit]

The geometric series 1/2 − 1/4 + 1/8 − 1/16 + ⋯ sums to 1/3.

The alternating harmonic series has a finite sum but the harmonic series does not.

The Mercator series provides an analytic expression of the natural logarithm:

$${\displaystyle \sum _{�=1}^{\mathrm{\infty }}\frac{(-1{)}^{�+1}}{�}{�}^{�}=\ln (1+�).}$$

The functions sine and cosine used in trigonometry can be defined as alternating series in calculus even though they are introduced in elementary algebra as the ratio of sides of a right triangle. In fact,

$${\displaystyle \sin �=\sum _{�=0}^{\mathrm{\infty }}(-1{)}^{�}\frac{{�}^{2�+1}}{(2�+1)!},}$$

and

$${\displaystyle \cos �=\sum _{�=0}^{\mathrm{\infty }}(-1{)}^{�}\frac{{�}^{2�}}{(2�)!}.}$$

When the alternating factor (–1)ⁿ is removed from these series one obtains the hyperbolic functions sinh and cosh used in calculus.

Proof of Alternating Series Test

Without loss of generality we can assume that the series starts at $n=1$. If not we could modify the proof below to meet the new starting place or we could do an index shift to get the series to start at $n=1$.

Also note that the assumption here is that we have $a_n={\left(-1\right)}^{n+1}{b}_n$. To get the proof for $a_n={\left(-1\right)}^n{b}_n$ we only need to make minor modifications of the proof and so will not give that proof.

Finally, in the examples all we really needed was for the $b_n$ to be positive and decreasing eventually but for this proof to work we really do need them to be positive and decreasing for all $n$.

First, notice that because the terms of the sequence are decreasing for any two successive terms we can say,

$$b_n-b_{n+1}\ge 0$$

Now, let’s take a look at the even partial sums.

$$\begin{array}{cccc}{s}_2& =b_1-b_2\ge 0& & \\ s_4& =b_1-b_2+b_3-b_4=s_2+b_3-b_4\ge s_2& & \text{because}{b}_3-b_4\ge 0\\ s_6& =s_4+b_5-b_6\ge s_4& & \text{because}{b}_5-b_6\ge 0\\ & ⋮& & \\ s_{2n}& =s_{2n-2}+b_{2n-1}-b_{2n}\ge s_{2n-2}& & \text{because}{b}_{2n-1}-b_{2n}\ge 0\end{array}$$

So, $\left\{s_{2n}\right\}$ is an increasing sequence.

Next, we can also write the general term as,

$$\begin{array}{cc}{s}_{2n}& =b_1-b_2+b_3-b_4+b_5+\cdots -b_{2n-2}+b_{2n-1}-b_{2n}\\ & =b_1-\left(b_2-b_3\right)-\left(b_4-b_5\right)+\cdots -\left(b_{2n-2}-b_{2n-1}\right)-b_{2n}\end{array}$$

Each of the quantities in parenthesis are positive and by assumption we know that $b_{2n}$ is also positive. So, this tells us that $s_{2n}\le b_1$ for all $n$.

We now know that $\left\{s_{2n}\right\}$ is an increasing sequence that is bounded above and so we know that it must also converge. So, let’s assume that its limit is $s$ or,

$$\underset{n\to \infty }{lim}{s}_{2n}=s$$

Next, we can quickly determine the limit of the sequence of odd partial sums, $\left\{s_{2n+1}\right\}$, as follows,

$$\underset{n\to \infty }{lim}{s}_{2n+1}=\underset{n\to \infty }{lim}\left(s_{2n}+b_{2n+1}\right)=\underset{n\to \infty }{lim}{s}_{2n}+\underset{n\to \infty }{lim}{b}_{2n+1}=s+0=s$$

So, we now know that both $\left\{s_{2n}\right\}$ and $\left\{s_{2n+1}\right\}$ are convergent sequences and they both have the same limit and so we also know that $\left\{s_n\right\}$ is a convergent sequence with a limit of $s$. This in turn tells us that $\sum a_n$ is convergent.

EXAMPLE  1  Determine if the following series is convergent or divergent.

$$$ \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}$$$

ANSWERE

First, identify the $b_n$ for the test.

$$\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}=\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{1}{n}{b}_n=\frac{1}{n}$$

Now, all that we need to do is run through the two conditions in the test.

$$\underset{n\to \infty }{lim}{b}_n=\underset{n\to \infty }{lim}\frac{1}{n}=0$$$$b_n=\frac{1}{n}>\frac{1}{n+1}=b_{n+1}$$

Both conditions are met and so by the Alternating Series Test the series must converge.

Example 2  Determine if the following series is convergent or divergent.

$$\sum _{n=2}^{\infty }\frac{\cos \left(n\pi \right)}{\sqrt{n}}$$

ANSWERE

The point of this problem is really just to acknowledge that it is in fact an alternating series. To see this we need to acknowledge that,

$$\cos \left(n\pi \right)={\left(-1\right)}^n$$

If you aren’t sure of this you can easily convince yourself that this is correct by plugging in a few values of $n$ and checking.

So the series is really,

$$\sum _{n=2}^{\infty }\frac{\cos \left(n\pi \right)}{\sqrt{n}}=\sum _{n=2}^{\infty }\frac{{\left(-1\right)}^n}{\sqrt{n}}\Rightarrow b_n=\frac{1}{\sqrt{n}}$$

Checking the two condition gives,

$$\underset{n\to \infty }{lim}{b}_n=\underset{n\to \infty }{lim}\frac{1}{\sqrt{n}}=0$$$$b_n=\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n+1}}=b_{n+1}$$

The two conditions of the test are met and so by the Alternating Series Test the series is convergent.

Example 3 Determine if the following series is convergent or divergent.

$$\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n{n}^2}{n^2+5}$$

ANSWERE

First, identify the $b_n$ for the test.

$$\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n{n}^2}{n^2+5}=\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{n^2}{n^2+5}\Rightarrow b_n=\frac{n^2}{n^2+5}$$

Let’s check the conditions.

$$\underset{n\to \infty }{lim}{b}_n=\underset{n\to \infty }{lim}\frac{n^2}{n^2+5}=1\ne 0$$

So, the first condition isn’t met and so there is no reason to check the second. In these cases where the first condition isn’t met it is usually best to use the divergence test.

So, the divergence test requires us to compute the following limit.

$$\underset{n\to \infty }{lim}\frac{{\left(-1\right)}^n{n}^2}{n^2+5}$$

This limit can be somewhat tricky to evaluate. For a second let’s consider the following,

$$\underset{n\to \infty }{lim}\frac{{\left(-1\right)}^n{n}^2}{n^2+5}=\left(\underset{n\to \infty }{lim}{\left(-1\right)}^n\right)\left(\underset{n\to \infty }{lim}\frac{n^2}{n^2+5}\right)$$

Splitting this limit like this can’t be done because this operation requires that both limits exist and while the second one does the first clearly does not.

So, let’s start with,

$$\underset{n\to \infty }{lim}\frac{{\left(-1\right)}^n{n}^2}{n^2+5}=\underset{n\to \infty }{lim}\left[{\left(-1\right)}^n\frac{n^2}{n^2+5}\right]$$

Now, the second part of this clearly is going to 1 as $n\to \infty$ while the first part just alternates between 1 and -1. So, as $n\to \infty$ 

In order for limits to exist we know that the terms need to settle down to a single number and since these clearly don’t this limit doesn’t exist and so by the Divergence Test this series diverges.

for more better understanding refer this video