Series

Example 1: Series Notation

The series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) represents the sum:

This particular series is known as the "Basel problem" and converges to \(\frac{\pi^2}{6}\).

Example 2: Partial Sums

Find the first few partial sums of the series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\).

Solution:

First, let's find the partial sums:

Now, let's find a simpler form for the general term using partial fractions:

Multiplying both sides by \(n(n+1)\):

Setting \(n=0\): \(1 = A\)

Setting \(n=1\): \(1 = A(2) + B = 2 + B\), so \(B = -1\)

Therefore:

Using our partial fraction decomposition, we can find a pattern:

This is a telescoping series where most terms cancel out, leaving only the first and last terms.

Example 3: Convergence of a Series

Determine whether the series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) converges or diverges. If it converges, find its sum.

Solution:

From Example 2, we found that the partial sum \(S_n = 1 - \frac{1}{n+1}\).

To determine if the series converges, we find the limit of the partial sums:

Therefore, the series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) converges to 1.

Example 13: Choosing the Right Test

Determine whether each of the following series converges or diverges. Use the most appropriate test for each.

(a) \(\sum_{n=1}^{\infty} \frac{3^n + n}{4^n}\)

Solution:

Let's try the ratio test:

Since \(\rho = \frac{3}{4} < 1\), the series \(\sum_{n=1}^{\infty} \frac{3^n + n}{4^n}\) converges by the ratio test.

(b) \(\sum_{n=2}^{\infty} \frac{1}{n \ln n}\)

Solution:

We apply the integral test. Let \(f(x) = \frac{1}{x \ln x}\) for \(x \geq 2\). We first verify the three hypotheses.

Continuity: Since \(x > 0\) and \(\ln x > 0\) for all \(x \geq 2\), the function \(f(x) = \frac{1}{x \ln x}\) is continuous on \([2, \infty)\).

Positivity: Since both \(x > 0\) and \(\ln x > 0\) for \(x \geq 2\), we have \(f(x) = \frac{1}{x \ln x} > 0\) on \([2, \infty)\).

Decreasing: We compute \(f'(x)\):

Since \(\ln x + 1 > 0\) for \(x \geq 2\), we have \(f'(x) < 0\) on \([2, \infty)\), so \(f\) is decreasing there.

Also, \(f(n) = \frac{1}{n \ln n} = a_n\) for all integers \(n \geq 2\). Since all hypotheses are satisfied, we evaluate the improper integral:

Let \(u = \ln x\), then \(du = \frac{1}{x} \, dx\).

As \(b \to \infty\), \(\ln(\ln b) \to \infty\), so the integral diverges. Therefore, the series \(\sum_{n=2}^{\infty} \frac{1}{n \ln n}\) diverges by the integral test.

(c) \(\sum_{n=1}^{\infty} (-1)^n \frac{\ln n}{n}\)

Solution:

This is an alternating series, so let's check if it satisfies the conditions of the alternating series test.

Let \(b_n = \frac{\ln n}{n}\).

1. \(b_n = \frac{\ln n}{n} > 0\) for all \(n \geq 2\). ✓
2. To check if \(b_n\) is decreasing, we can find the derivative of \(f(x) = \frac{\ln x}{x}\):
   \begin{align} f'(x) &= \frac{1}{x} \cdot \frac{1}{x} - \frac{\ln x}{x^2} \\ &= \frac{1 - \ln x}{x^2} \end{align}
   This is negative for \(x > e\), so \(b_n\) is decreasing for \(n \geq 3\). ✓
3. \(\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{\ln n}{n}\). Using L'Hôpital's rule for the corresponding real-valued function:
   \begin{align} \lim_{x \to \infty} \frac{\ln x}{x} &= \lim_{x \to \infty} \frac{\frac{1}{x}}{1} \\ &= \lim_{x \to \infty} \frac{1}{x} \\ &= 0 \end{align}
   So it follows that \(\lim_{n \to \infty} b_n = 0\). ✓

Since all three conditions are satisfied (after the first few terms), the alternating series \(\sum_{n=3}^{\infty} (-1)^n \frac{\ln n}{n}\) converges by the alternating series test and hence \(\sum_{n=1}^{\infty} (-1)^n \frac{\ln n}{n}\) also converges.