Question

Test the following series for convergence. $$\sum_{n=0}^{\infty} \frac{(-1)^{n} n}{1+n^{2}}$$

Short answer

The series \( \frac{(-1)^n n}{1+n^2} \) converges by the Alternating Series Test.

Step-by-step solution

Identify the series type

Recognize that the series \( \frac{(-1)^n n}{1+n^2} \) is an alternating series due to the factor \((-1)^n\). This indicates it may be evaluated using the Alternating Series Test (Leibniz's Test).

Define the sequence terms

Let \( a_n = \frac{n}{1+n^2} \). The series can then be written as \( \frac{(-1)^n n}{1+n^2} = (-1)^n \times a_n \).

Check the decreasing nature

To apply the Alternating Series Test, \(a_n\) must be monotonically decreasing for all \(n\). Observe the behaviour of \( a_n = \frac{n}{1+n^2} \) as \( n \) increases. Calculate \( a_{n+1} \) and compare it with \( a_n \).

Take the derivative of the sequence term

Another way to show that \(a_n\) is decreasing is by taking its derivative. Compute \( a'_n \) to show that it is negative. Let \( f(n) = \frac{n}{1+n^2} \). Using the quotient rule, \[ f'(n) = \frac{(1+n^2) - n \times 2n}{(1+n^2)^2} = \frac{1+n^2-2n^2}{(1+n^2)^2} = \frac{1-n^2}{(1+n^2)^2} \]. For all \(n eq 1\), \( 1-n^2 < 0 \) for \( n > 1 \), thus \( f'(n) < 0 \), indicating that \(a_n\) is decreasing.

Check the limit of sequence terms

Finally, the second condition of the Alternating Series Test requires that \( \text{lim}_{n \to \infty} a_n = 0 \). Evaluate the limit: \[ \text{lim}_{n \to \infty} \frac{n}{1+n^2} = \text{lim}_{n \to \infty} \frac{1/n}{1/n^2 + 1} = 0 \].

Conclude the convergence

Since \(a_n\) is monotonically decreasing and \( \text{lim}_{n \to \infty} a_n = 0 \), both conditions of the Alternating Series Test are satisfied. Therefore, the series \( \frac{(-1)^n n}{1+n^2} \) converges.

Key concepts

Alternating Series Test

When dealing with series, one useful tool is the Alternating Series Test (also known as Leibniz's Test). This test helps determine if an alternating series converges. An alternating series is a series where the signs of the terms alternate, like in the sequence given by \( (-1)^n \). To apply this test, you need to confirm two criteria:

• The sequence terms \( a_n \) must be monotonically decreasing.
• The limit of the sequence \( a_n \) as \( n \to \infty \) must be zero.

For example, in the series \( \sum_{n=0}^{\infty} \frac{(-1)^n n}{1+n^2} \), the factor \((-1)^n\) makes it an alternating series, which is our first clue that the Alternating Series Test is applicable here.

Monotonically Decreasing Sequence

A sequence is monotonically decreasing if each term is less than or equal to the previous term. For the sequence \( a_n = \frac{n}{1+n^2} \), we need to verify this property to apply the Alternating Series Test.
To do this, we examine the sequence's behavior as \( n \) increases. Alternatively, we can use calculus to take the derivative \( a'_n \). If \( a'_n < 0 \) for all \ n > 1 \, then the sequence is monotonically decreasing.

• First, compute \( a_{n+1} = \frac{n+1}{1+(n+1)^2} \).
• Then, show that \ a_{n+1} < a_n \, which mathematically confirms that the sequence is decreasing.
• You can also use derivatives to ease this process.

Limit of Sequence Terms

The second condition from the Alternating Series Test is that the limit of the sequence terms \(a_n \) as \ n \to \infty \, must be zero.
In our example, this means we examine:
\ \lim_{n \to \infty} \frac{n}{1+n^2} \.
To find this limit:

• Observe that as \ n \ grows large, \ n^2 \ will dominate over \ n\.
• Simplify: \ \frac{n}{1+n^2} \to \frac{1/n}{1/n^2 + 1} = \frac{1}{n+n}\. This tends towards zero as \ n \ goes to infinity.

Since the limit is zero, this condition is satisfied for our sequence.

Quotient Rule

Sometimes, you need to compute the derivative of a function when checking for its monotonic nature. The quotient rule is handy for such a task.
The quotient rule states that for a function \ f(x) = \frac{u(x)}{v(x)} \, the derivative \ f'(x) \ is given by:

\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \]
For our sequence \( a_n = \frac{n}{1+n^2} \, where \ u(n) = \ n \ and \ v(n) = 1 + n^2 \), applying the quotient rule yields:
\ f'(n) = \frac{(1+n^2) - n \times 2n}{(1+n^2)^2} = \frac{1 - n^2}{(1+n^2)^2} \.
Notice that for \ n > 1 \, \ 1 - n^2 \ is negative, meaning \ f'(n) < 0 \. This confirms that \ a_n \ is monotonically decreasing, fulfilling another requirement for the Alternating Series Test.