Infinite Series & Convergence
Infinite Series & Convergence
$$\sum_{k=1}^{\infty} a_k = \lim_{n\to\infty} S_n$$
If the limit exists and is finite, the series converges; otherwise it diverges.
If $|r| < 1$:
$$\sum_{k=0}^{\infty} a_1 r^k = \frac{a_1}{1 - r}$$
If $|r| \geq 1$, the series diverges.
If $\lim_{n\to\infty} a_n \neq 0$, then $\sum a_n$ diverges.
Warning: If $\lim a_n = 0$, the series may still diverge (e.g., harmonic series).
A series where consecutive terms cancel, leaving only the first and last pieces. Write partial fractions and cancel.
Find the sum: $\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k$.
$a_1 = 1$, $r = 1/2$. Since $|r| < 1$:
$$S = \frac{1}{1 - 1/2} = 2$$
Does $\sum_{k=1}^{\infty} \frac{k}{k+1}$ converge?
$\lim_{k\to\infty} \frac{k}{k+1} = 1 \neq 0$. By the Divergence Test, the series diverges.
Write the repeating decimal $0.333\ldots$ as a fraction.
$0.333\ldots = \sum_{k=1}^{\infty} 3 \cdot (0.1)^k = \frac{0.3}{1 - 0.1} = \frac{0.3}{0.9} = \frac{1}{3}$
Practice Problems
Show Answer Key
1. $\frac{1}{1-1/3} = 3/2$
2. $\frac{1}{1-3/4} = 4$
3. No — it is the harmonic series (diverges)
4. $\lim (-1)^k$ does not exist (alternates), so diverges
5. $\frac{5 \cdot 0.2}{1 - 0.2} = \frac{1}{0.8} = 1.25$
6. $\frac{0.6}{1-0.1} = \frac{0.6}{0.9} = 2/3$
7. No, $|r| = 2 \geq 1$
8. Partial fractions: $\frac{1}{k} - \frac{1}{k+1}$. Telescopes to $1$.
9. $\frac{10}{1-(-0.5)} = \frac{10}{1.5} = 20/3$
10. $\frac{3/10}{1-1/10} = \frac{3/10}{9/10} = 1/3$
11. If $\lim a_n \neq 0$, the series diverges
12. No, it diverges