Geometric & Telescoping Series

$\dfrac{5 \cdot 2^{n-1}}{3^{n+1}} = \dfrac{5}{2 \cdot 3} \cdot \dfrac{2^{n}}{3^{n}} = \dfrac{5}{6}\left(\dfrac{2}{3}\right)^{n}$.

$r = \tfrac{2}{3}$ and $\left|\tfrac{2}{3}\right| < 1$, so it converges.

At $n = 2$: $\dfrac{5}{6} \cdot \dfrac{4}{9} = \dfrac{10}{27}$.

$\dfrac{10/27}{1 - 2/3} = \dfrac{10}{27} \cdot 3 = \dfrac{10}{9}$.

$\dfrac{2^{n} + (-1)^{n}}{5^{n}} = \left(\dfrac{2}{5}\right)^{n} + \left(-\dfrac{1}{5}\right)^{n}$: two geometric series.

$r = \tfrac{2}{5}$ and $r = -\tfrac{1}{5}$: both have $|r| < 1$.

At $n = 0$ both first terms are $1$.

$\dfrac{1}{1 - 2/5} + \dfrac{1}{1 + 1/5} = \dfrac{5}{3} + \dfrac{5}{6} = \dfrac{5}{2}$.

$\dfrac{3^{2n}}{8^{n}} = \dfrac{9^{n}}{8^{n}} = \left(\dfrac{9}{8}\right)^{n}$.

$r = \tfrac{9}{8} \geq 1$: the terms grow, so the series diverges.

Not needed: the series diverges.

No sum to compute.

Nothing to split off: the whole decimal repeats.

$0.4545\ldots = \dfrac{45}{100} + \dfrac{45}{100^{2}} + \cdots$, so $r = \dfrac{1}{100}$.

$\dfrac{45/100}{1 - 1/100} = \dfrac{45}{99}$.

$\dfrac{45}{99} = \dfrac{5}{11}$.

$1.2\overline{7} = \dfrac{12}{10} + 0.0777\ldots$

$0.0777\ldots = \dfrac{7}{100} + \dfrac{7}{1000} + \cdots$, so $r = \dfrac{1}{10}$.

$\dfrac{7/100}{1 - 1/10} = \dfrac{7}{90}$.

$\dfrac{12}{10} + \dfrac{7}{90} = \dfrac{108}{90} + \dfrac{7}{90} = \dfrac{115}{90} = \dfrac{23}{18}$.

$3.\overline{108} = 3 + 0.108108\ldots$

$0.108108\ldots = \dfrac{108}{1000} + \dfrac{108}{1000^{2}} + \cdots$, so $r = \dfrac{1}{1000}$.

$\dfrac{108/1000}{1 - 1/1000} = \dfrac{108}{999} = \dfrac{4}{37}$.

$3 + \dfrac{4}{37} = \dfrac{115}{37}$.

$\dfrac{2}{n(n+2)} = \dfrac{1}{n} - \dfrac{1}{n+2}$.

$S_N = \left(1 - \tfrac{1}{3}\right) + \left(\tfrac{1}{2} - \tfrac{1}{4}\right) + \cdots + \left(\tfrac{1}{N} - \tfrac{1}{N+2}\right) = 1 + \tfrac{1}{2} - \tfrac{1}{N+1} - \tfrac{1}{N+2}$.

$S_N \to 1 + \tfrac{1}{2} = \tfrac{3}{2}$.

Already a difference: $b_{n+1} - b_n$, with $b_n = \arctan n$.

$S_N = (\arctan 2 - \arctan 1) + (\arctan 3 - \arctan 2) + \cdots + \big(\arctan(N+1) - \arctan N\big) = \arctan(N+1) - \arctan 1$.

$S_N \to \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$: a finite limit, so it converges.

$\dfrac{n^{2}}{(n-1)(n+1)} = \dfrac{n}{n-1} \cdot \dfrac{n}{n+1}$, so the term is $\log\dfrac{n}{n-1} - \log\dfrac{n+1}{n}$.

With $b_n = \log\dfrac{n}{n-1}$, everything between cancels: $S_N = b_2 - b_{N+1} = \log 2 - \log\dfrac{N+1}{N}$.

$\log\dfrac{N+1}{N} \to \log 1 = 0$, so $S_N \to \log 2$.

$a_1 = S_1 = \dfrac{2}{4} = \dfrac{1}{2}$.

$a_n = \dfrac{2n}{3n + 1} - \dfrac{2(n-1)}{3n - 2} = \dfrac{2}{(3n + 1)(3n - 2)}$. At $n = 1$ this gives $\tfrac{2}{4} = \tfrac{1}{2}$, so it holds for every $n$.

$\lim_{N\to\infty} \dfrac{2N}{3N + 1} = \dfrac{2}{3}$.

$a_1 = S_1 = 3 - \tfrac{1}{2} = \tfrac{5}{2}$.

$a_n = \left(3 - \dfrac{1}{2^{n}}\right) - \left(3 - \dfrac{1}{2^{n-1}}\right) = \dfrac{1}{2^{n}}$ for $n \geq 2$. At $n = 1$ this gives $\tfrac{1}{2} \neq \tfrac{5}{2}$, so $a_1$ is special.

$S_N \to 3 - 0 = 3$.

$a_1 = S_1 = \log 2$.

$a_n = \log(n + 1) - \log n = \log\dfrac{n + 1}{n}$; at $n = 1$ this is $\log 2$, so it holds for every $n$.

$S_N = \log(N + 1) \to \infty$, so the series diverges, even though $a_n \to 0$.