Alternating Series

Going back to the Alternating Harmonic Series, you already know that the first condition of the Alternating Series Test is satisfied because

$a_n=\frac{1}{n}>0$.

Condition 2: Since

$a_n=\frac{1}{n}$ and $a_{n+1}=\frac{1}{n+1}$

you know that $a_n>\hspace{0.17em}{a}_{n+1}$ because$n<n+1$, which implies that

$\frac{1}{n}>\hspace{0.17em}\frac{1}{n+1}$,

so the second condition is satisfied.

Condition 3: You also know that

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{1}{n}=0$,

so the third condition is satisfied.

That means you can apply the Alternating Series Test to say that the Alternating Harmonic Series converges.

Remember that the Harmonic Series diverges, so sometimes changing something from a regular series to an alternating series can change it from diverging to converging.

You know that the P-series

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

converges because $p=2$.

What can you say about the Alternating P-series

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

Answer: Before applying the Alternating Series Test you need to be sure that all of the conditions are satisfied.

Condition 1: Here

$a_n=\frac{1}{n^2}>0$

so the first condition is satisfied.

Condition 2: You know that

$a_n=\frac{1}{n^2}$

and

$a_{n+1}=\frac{1}{{\left(n+1\right)}^2}=\frac{1}{n^2+2n+1}<\frac{1}{n^2}=a_n$,

so the second condition is satisfied.

Condition 3: Looking at the limit,

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{1}{n^2}=0$,

so the third condition is satisfied.

That means the Alternating Series Test tells you that the series

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

converges.

Can you use the Alternating Series Test to tell if the series

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

converges?

Answer:

Let's check the conditions for the Alternating Series Test.

Condition 1: For this series

$a_n=\frac{1}{\sqrt{n}}>0$,

so the first condition is satisfied.

Condition 2: Here

$a_n=\frac{1}{\sqrt{n}}$ and $a_{n+1}=\frac{1}{\sqrt{n+1}}$.

You may not have a good intuition of how $a_n$ and $a_{n+1}$ compare to each other, so let's try out a value for $n$ and see what happens. If $n=10$, then

$a_{10}=\frac{1}{\sqrt{10}}\approx 0.3162$, and $a_{11}=\frac{1}{\sqrt{11}}\approx 0.3015$,

which means that $a_{10}>\hspace{0.17em}{a}_{11}$. This is the direction you would hope the inequality would go. In fact, since $n<n+1$, you know that $\sqrt{n}<\sqrt{n+1}$ which implies

$\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n+1}}$,

so in general $a_n>a_{n+1}$. Now you know that the second condition holds.

Condition 3: Taking a look at the limit,

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n}}=0$,

so the third condition holds as well.

Now by the Alternating Series Test, you know that the series converges.

Similar to the example earlier, you can show that the partial sums for the series

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

diverges, which means the series diverges. What happens if you try and use the Alternating Series Test?

Answer:

First, let's look at the sequence of partial sums for this series. You know that

$\begin{array}{ccc}{s}_1& =& 1\\ s_2& =& 1-1=0\\ s_3& =& 0+\hspace{0.17em}1=1\\ s_4& =& 1-1=0\\ & ⋮& \end{array}$

so in fact the sequence of partial sums diverges, which by definition means the series diverges.

To apply the Alternating Series Test you need all three conditions satisfied. For this series, $a_n=1$ for any $n$, which means the first condition is satisfied. But when you check the second condition you need that $a_n>a_{n+1}$, or in other words, that $1>1$ which isn't true. It is even worse when you look at the limit in the third condition since

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }1=1$,

which is definitely not zero. So even though you know by looking at the partial sums that this series diverges, you can't use the Alternating Series Test to say anything about it.

Look at the series

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

This series diverges, which means that one or more of the three conditions of the Alternating Series Test fails. Which condition fails?

Answer:

Here

$a_n=\frac{n}{2n-1}$,

and you can see that the first condition holds.

Condition 2: To check condition 2, showing that $a_n>\hspace{0.17em}{a}_{n+1}$ is the same as showing that $a_n-a_{n+1}>0$. So

id="2905224" role="math" $$\begin{gathered} a_n-a_{n+1}=\frac{n}{2n-1}-\frac{n+1}{2(n+1)-1} \\ =\frac{1}{\left(2n-1\right)\left(2n+\hspace{0.17em}1\right)} \\ =\frac{1}{4n^2-1} \\ >\frac{1}{4n^2} \\ >0 \end{gathered}$$

and the second condition holds.

Condition 3: Looking at the limit,

$\underset{n\to \infty }{\lim }\frac{n}{2n-1}=\frac{1}{2}$,

which is definitely not zero. That means condition 3 of the Alternating Series Test fails.

Let's take a look at the Alternating Harmonic series. You already know that the terms of the Alternating Series Bound theorem are satisfied from an earlier example. That means you are safe to apply the Alternating Series Bound theorem.

Suppose you add up the first 10 terms. How far off is the partial sum from the actual answer?

Answer: If you do the calculation,

$s_{10}=\frac{1267}{2520}\approx 0.6456$,

so using the Alternating Series Bound theorem the truncation error for $s_{10}$ is less than

id="2934012" role="math" $a_{11}=\frac{1}{11}$.

The fact that the term ${\left(-1\right)}^{11}$ is negative tells you that this is an underestimate rather than an overestimate.