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education/math/MATH1220 (calc II)/Sequences.md
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@ -40,7 +40,7 @@ Remember, L'Hospital's rule states that:
# Series # Series
Vocabulary: A **series** is another name for a sum of numbers. Vocabulary: A **series** is another name for a sum of numbers.
## Properties ## The Limit Test
You can break a series into *partial sums*: You can break a series into *partial sums*:
$$\sum_{n=1}^\infty a_n = a_1 + 1_2 + a_3 + ...$$ $$\sum_{n=1}^\infty a_n = a_1 + 1_2 + a_3 + ...$$
Given the above series, we can define the following: Given the above series, we can define the following:
@ -52,10 +52,16 @@ Given the above series, we can define the following:
and say that the sum converges to $L$. and say that the sum converges to $L$.
## Examples ## Examples
> Prove that $\sum_{n = 1}^\infty a_n = L$ > Prove that $\sum_{n = 1}^\infty \frac{1}{2^n} = 1$
- $S_1 = \frac{1}{2}$ - $S_1 = \frac{1}{2}$
- $S_2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ - $S_2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$
- $S_3 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$ - $S_3 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$
- $S_n = \frac{2^n - 1}{2^n}$ - $S_n = \frac{2^n - 1}{2^n}$
So: So:
$$ \sum_{n=1}^\infty \frac{1}{2^n} = \lim_{n \to \infty}S_n = \lim_{n \to \infty} (1 - \frac{}{}$$ $$ \sum_{n=1}^\infty \frac{1}{2^n} = \lim_{n \to \infty}S_n = \lim_{n \to \infty} (1 - \frac{1}{2^n}) = 1$$
> Using the limit test, determine whether the series $\sum_{n = 1}^\infty n$ converges or diverges
- $S_1 = 1$
- $S_2 = 1 + 2 = 3$
- $S_n = \frac{n(n+1)}{2}$
So: