Geometric Series for MATH 141
Exam Relevance for MATH 141
Geometric series are foundational for MATH 141 series work. Know the sum formula and recognize disguised geometric series.
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What is a Geometric Series?
A geometric series is a series where each term is a constant multiple of the previous term. It's the most important series in calculus because it's one of the few infinite series we can sum exactly.
$$a + ar + ar^2 + ar^3 + \cdots = \sum_{n=0}^{\infty} ar^n$$
- $a$ = first term
- $r$ = common ratio (each term is $r$ times the previous)
The Geometric Series Formula
$$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \quad \text{if } |r| < 1$$
Convergence: The series converges only when $|r| < 1$.
Divergence: The series diverges when $|r| \geq 1$.
Why $|r| < 1$?
- If $|r| < 1$: Each term is smaller than the last, so the sum can settle to a finite value
- If $|r| = 1$: Terms don't shrink (e.g., $1 + 1 + 1 + \cdots$) — diverges
- If $|r| > 1$: Terms grow larger and larger — diverges
Identifying Geometric Series
Problem: Identify the first term $a$ and common ratio $r$ for $\displaystyle\sum_{n=0}^{\infty} \frac{3}{5^n}$.
Rewrite to match the form $ar^n$:
$$\frac{3}{5^n} = 3 \cdot \frac{1}{5^n} = 3 \cdot \left(\frac{1}{5}\right)^n$$
Comparing to $ar^n$:
$$\boxed{a = 3, \quad r = \frac{1}{5}}$$
Problem: Identify $a$ and $r$ for $\displaystyle\sum_{n=1}^{\infty} 2^{-n}$.
First, note this starts at $n = 1$, not $n = 0$.
$$2^{-n} = \frac{1}{2^n} = \left(\frac{1}{2}\right)^n$$
The first term (when $n = 1$): $a = \left(\frac{1}{2}\right)^1 = \frac{1}{2}$
The ratio: $r = \frac{1}{2}$
$$\boxed{a = \frac{1}{2}, \quad r = \frac{1}{2}}$$
Finding the Sum
Problem: Find the sum: $\displaystyle\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n$.
Identify: $a = 1$ (when $n=0$: $\left(\frac{2}{3}\right)^0 = 1$), $r = \frac{2}{3}$
Check convergence: $|r| = \frac{2}{3} < 1$ ✓
Apply the formula:
$$\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n = \frac{a}{1-r} = \frac{1}{1 - \frac{2}{3}} = \frac{1}{\frac{1}{3}} = \boxed{3}$$
Problem: Find the sum: $\displaystyle\sum_{n=0}^{\infty} \frac{4}{7^n}$.
$$\frac{4}{7^n} = 4 \cdot \left(\frac{1}{7}\right)^n$$
So $a = 4$, $r = \frac{1}{7}$.
Check: $|r| = \frac{1}{7} < 1$ ✓
$$\sum_{n=0}^{\infty} \frac{4}{7^n} = \frac{4}{1 - \frac{1}{7}} = \frac{4}{\frac{6}{7}} = 4 \cdot \frac{7}{6} = \boxed{\frac{14}{3}}$$
Problem: Find the sum: $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{4^n}$.
$$\frac{(-1)^n}{4^n} = \left(\frac{-1}{4}\right)^n$$
So $a = 1$, $r = -\frac{1}{4}$.
Check: $|r| = \frac{1}{4} < 1$ ✓
$$\sum_{n=0}^{\infty} \left(-\frac{1}{4}\right)^n = \frac{1}{1 - \left(-\frac{1}{4}\right)} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \boxed{\frac{4}{5}}$$
Problem: Find the sum: $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$.
Method 1: Adjust the first term
When $n = 1$: first term is $\left(\frac{1}{2}\right)^1 = \frac{1}{2}$
So $a = \frac{1}{2}$, $r = \frac{1}{2}$.
$$\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = \boxed{1}$$
Method 2: Subtract the $n=0$ term
$$\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n - \left(\frac{1}{2}\right)^0 = \frac{1}{1-\frac{1}{2}} - 1 = 2 - 1 = 1 \quad ✓$$
Problem: Find the sum: $\displaystyle\sum_{n=2}^{\infty} 3 \cdot \left(\frac{1}{5}\right)^n$.
Method: Find the first term at $n = 2$
First term (when $n = 2$): $a = 3 \cdot \left(\frac{1}{5}\right)^2 = 3 \cdot \frac{1}{25} = \frac{3}{25}$
Ratio: $r = \frac{1}{5}$
$$\sum_{n=2}^{\infty} 3 \cdot \left(\frac{1}{5}\right)^n = \frac{\frac{3}{25}}{1 - \frac{1}{5}} = \frac{\frac{3}{25}}{\frac{4}{5}} = \frac{3}{25} \cdot \frac{5}{4} = \boxed{\frac{3}{20}}$$
Recognizing Geometric Series in Disguise
Problem: Find the sum: $\displaystyle\sum_{n=0}^{\infty} \frac{2^n}{3^{n+1}}$.
Rewrite:
$$\frac{2^n}{3^{n+1}} = \frac{2^n}{3 \cdot 3^n} = \frac{1}{3} \cdot \frac{2^n}{3^n} = \frac{1}{3} \cdot \left(\frac{2}{3}\right)^n$$
Now it's geometric with $a = \frac{1}{3}$, $r = \frac{2}{3}$.
$$\sum_{n=0}^{\infty} \frac{2^n}{3^{n+1}} = \frac{\frac{1}{3}}{1 - \frac{2}{3}} = \frac{\frac{1}{3}}{\frac{1}{3}} = \boxed{1}$$
Problem: Find the sum: $\displaystyle\sum_{n=0}^{\infty} \frac{e^n}{\pi^n}$.
$$\frac{e^n}{\pi^n} = \left(\frac{e}{\pi}\right)^n$$
Since $e \approx 2.718$ and $\pi \approx 3.14$, we have $r = \frac{e}{\pi} < 1$ ✓
$$\sum_{n=0}^{\infty} \left(\frac{e}{\pi}\right)^n = \frac{1}{1 - \frac{e}{\pi}} = \frac{1}{\frac{\pi - e}{\pi}} = \boxed{\frac{\pi}{\pi - e}}$$
Divergent Geometric Series
Problem: Determine if $\displaystyle\sum_{n=0}^{\infty} \left(\frac{5}{3}\right)^n$ converges or diverges.
Here $r = \frac{5}{3}$.
Since $|r| = \frac{5}{3} > 1$:
$$\boxed{\text{The series diverges}}$$
We cannot use the formula $\frac{a}{1-r}$ when $|r| \geq 1$.
Problem: Determine if $\displaystyle\sum_{n=0}^{\infty} (-1)^n$ converges or diverges.
Here $r = -1$, so $|r| = 1$.
Since $|r| = 1$ (not less than 1):
$$\boxed{\text{The series diverges}}$$
The partial sums alternate: $1, 0, 1, 0, 1, 0, \ldots$ — never settling on a value.
Repeating Decimals as Geometric Series
One beautiful application: every repeating decimal is a geometric series!
Problem: Express $0.333\ldots$ as a fraction.
$$0.333\ldots = 0.3 + 0.03 + 0.003 + \cdots$$
$$= \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots$$
$$= \frac{3}{10}\left(1 + \frac{1}{10} + \frac{1}{100} + \cdots\right)$$
$$= \frac{3}{10} \cdot \frac{1}{1 - \frac{1}{10}} = \frac{3}{10} \cdot \frac{10}{9} = \boxed{\frac{1}{3}}$$
Problem: Express $0.\overline{27} = 0.272727\ldots$ as a fraction.
$$0.272727\ldots = \frac{27}{100} + \frac{27}{10000} + \frac{27}{1000000} + \cdots$$
$$= \frac{27}{100}\left(1 + \frac{1}{100} + \frac{1}{10000} + \cdots\right)$$
This is geometric with $a = 1$, $r = \frac{1}{100}$.
$$= \frac{27}{100} \cdot \frac{1}{1 - \frac{1}{100}} = \frac{27}{100} \cdot \frac{100}{99} = \boxed{\frac{27}{99} = \frac{3}{11}}$$
Finite Geometric Sums
For reference, the finite geometric sum formula is:
$$\sum_{n=0}^{N} ar^n = a \cdot \frac{1 - r^{N+1}}{1 - r} \quad (r \neq 1)$$
As $N \to \infty$, if $|r| < 1$, then $r^{N+1} \to 0$, giving us back $\frac{a}{1-r}$.
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting to check $|r| < 1$
Wrong: "$\displaystyle\sum_{n=0}^{\infty} 2^n = \frac{1}{1-2} = -1$"
Why it's wrong: When $r = 2$, we have $|r| > 1$, so the series diverges. The formula doesn't apply!
Correct: First check: $|r| = 2 > 1$, so the series diverges. The formula $\frac{a}{1-r}$ is meaningless here.
❌ Mistake: Wrong first term when $n$ doesn't start at 0
Wrong: For $\displaystyle\sum_{n=1}^{\infty} 3^{-n}$, using $a = 1$.
Why it's wrong: When $n = 1$, the first term is $3^{-1} = \frac{1}{3}$, not $3^0 = 1$.
Correct: Always find $a$ by plugging in the starting index. Here $a = 3^{-1} = \frac{1}{3}$.
❌ Mistake: Confusing $a$ and $r$
Wrong: For $\displaystyle\sum_{n=0}^{\infty} \frac{5}{2^n}$, saying $a = \frac{5}{2}$.
Why it's wrong: Rewrite as $5 \cdot \left(\frac{1}{2}\right)^n$. When $n = 0$, the first term is $5 \cdot 1 = 5$.
Correct: $a = 5$ (the first term), $r = \frac{1}{2}$ (what you multiply by each time).
❌ Mistake: Using the formula when $r = 1$
Wrong: "$\displaystyle\sum_{n=0}^{\infty} 1 = \frac{1}{1-1} = \frac{1}{0}$, which is undefined, so... it converges to undefined?"
Why it's wrong: Division by zero isn't "undefined convergence" — it means the formula doesn't apply. When $r = 1$, the series is $1 + 1 + 1 + \cdots$, which clearly diverges.
Correct: When $|r| \geq 1$, say "the series diverges" — don't try to use the formula.
Geometric Series Sum
The sum of an infinite geometric series. Only valid when |r| < 1! When |r| ≥ 1, the series diverges and this formula cannot be used.
Variables:
- $a$:
- the first term (when n=0)
- $r$:
- the common ratio
- $|r|$:
- absolute value of r (must be less than 1)
Geometric Series Convergence Criterion
A geometric series converges if and only if the absolute value of the common ratio is less than 1. When |r| ≥ 1, it diverges.
Variables:
- $r$:
- the common ratio
Finite Geometric Sum
The sum of the first N+1 terms of a geometric series. Works for any r ≠ 1. As N → ∞ with |r| < 1, this approaches a/(1-r).
Variables:
- $a$:
- the first term
- $r$:
- the common ratio
- $N$:
- the final index (sum has N+1 terms)
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