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education/math/MATH1220 (calc II)/Sequences.md
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@@ -82,4 +82,7 @@ $$ = \sum_{n=1}^\infty 35*(\frac{1}{7})^n*2^{n-1}$$
2. Pull a $7$ out of the bottom, making use of the fact that $(\frac{1}{7})^n = \frac{1}{7}(\frac{1}{7})^{n-1}$ 2. Pull a $7$ out of the bottom, making use of the fact that $(\frac{1}{7})^n = \frac{1}{7}(\frac{1}{7})^{n-1}$
$$ = \sum_{n=1}^\infty \frac{35}{7}(\frac{1}{7})^{n-1}* 2^{n-1} = \sum_{n = 1}^\infty \frac{35}{7}(\frac{2}{7})^{n-1} $$ $$ = \sum_{n=1}^\infty \frac{35}{7}(\frac{1}{7})^{n-1}* 2^{n-1} = \sum_{n = 1}^\infty \frac{35}{7}(\frac{2}{7})^{n-1} $$
3. This is now of the form $\sum_{n=1}^\infty ar^{n-1}$, so: 3. This is now of the form $\sum_{n=1}^\infty ar^{n-1}$, so:
$$\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) = \dfrac{\frac{35}{7}}{1}$$ $$\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) = \dfrac{\frac{35}{7}}{1-\frac{2}{7}}$$
# Divergence Test
If $\lim_{n \to \infty} a_n \ne 0$ then $\sum_{a}