For most series you can never write $S_N$ in closed form. Two families are exceptions, and they are the workhorses of the subject: geometric series (you met infinite geometric series in algebra) and telescoping series. For both, you can find the partial sums exactly.
Por qué funciona
Geometric. Let $S_N = a + ar + \cdots + ar^{N-1}$. Then $rS_N = ar + \cdots + ar^{N}$, and subtracting cancels everything in the middle: $S_N - rS_N = a - ar^{N}$, so $S_N = \frac{a(1 - r^{N})}{1 - r}$ for $r \ne 1$. If $|r| < 1$, $r^{N} \to 0$ and $S_N \to \frac{a}{1 - r}$. If $|r| \ge 1$, the terms do not go to 0 and the series diverges.
Telescoping. If each term is a difference $b_n - b_{n+1}$, the partial sum collapses: $S_N = b_1 - b_{N+1}$. The series converges exactly when $b_{N+1}$ has a limit.
Idea clave
$$\sum_{n=0}^{\infty} ar^{n} = \frac{a}{1 - r} = \frac{\text{first term}}{1 - \text{ratio}} \quad (|r| < 1);$$
it diverges for $|r| \ge 1$. A telescoping series $\sum (b_n - b_{n+1})$ has $S_N = b_1 - b_{N+1}$.
Ejemplo resuelto: Mind the first term
Find $\sum_{n=1}^{\infty} 3\left(\frac{2}{5}\right)^{n}$.
- Ratio $r = \frac{2}{5}$, and $|r| < 1$.
- First term ($n = 1$): $3\cdot\frac{2}{5} = \frac{6}{5}$.
- Sum: $\frac{6/5}{1 - 2/5} = \frac{6/5}{3/5} = 2$.
Respuesta: $2$
Error común
Using the coefficient instead of the first term. $\sum_{n=1}^{\infty} 3\left(\frac{2}{5}\right)^{n} \ne \frac{3}{1 - 2/5} = 5$, because the series starts at $n = 1$, so its first term is $\frac{6}{5}$, not 3. Check: write out $S_1$ and use it as the numerator.
Ejemplo resuelto: A repeating decimal
Write $0.\overline{36}$ as a fraction.
- $0.363636\ldots = \frac{36}{100} + \frac{36}{100^{2}} + \cdots$, geometric with first term $\frac{36}{100}$ and ratio $\frac{1}{100}$.
- Sum: $\frac{36/100}{1 - 1/100} = \frac{36}{99} = \frac{4}{11}$.
Respuesta: $\frac{4}{11}$
Ejemplo resuelto: Telescoping
Find $\sum_{n=1}^{\infty}\frac{1}{n(n + 2)}$.
- Partial fractions: $\frac{1}{n(n + 2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n + 2}\right)$.
- In $S_N$, each $-\frac{1}{n + 2}$ cancels the $+\frac{1}{n}$ two terms later. What survives: $\frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N + 1} - \frac{1}{N + 2}\right)$.
- As $N \to \infty$: $\frac{1}{2}\cdot\frac{3}{2} = \frac{3}{4}$.
Respuesta: $\frac{3}{4}$
Ejemplo resuelto: When the ratio is too big
What about $\sum_{n=0}^{\infty}\left(-\frac{3}{2}\right)^{n}$?
- Here $|r| = \frac{3}{2} \ge 1$, so the terms grow in size and the series diverges.
- Plugging into $\frac{a}{1 - r}$ would give $\frac{1}{1 + 3/2} = \frac{2}{5}$, a meaningless number: the formula only holds when $|r| < 1$.
Respuesta: It diverges.