Evaluate the partial sum S_n and the sum S = lim as n
Evaluate the partial sum $S_n$ and the sum $S = \lim_{n \to \infty} S_n$ for the series: $$ \sum_{n=1}^{\infty} q^n. $$
Explanation of Method
The given series is geometric, starting from $n=1$. A closed form for the partial sum $S_n$ can be found with the classic shift-and-subtract trick: multiply $S_n$ by $q$, then subtract the result from $S_n$ itself. Nearly every term cancels, leaving an expression that can be solved directly for $S_n$. The sum $S$ then follows by letting $n \to \infty$ and examining how the term $q^{n+1}$ behaves depending on the size of $q$.
- Form $S_n$ and $qS_n$, then subtract to isolate $S_n$.
- Take the limit of $S_n$ as $n \to \infty$, splitting into cases based on $|q|$.
Step 1 · Determine the partial sum $S_n$
The $n$-th partial sum is the sum of the first $n$ terms:
$$ S_n = q^1 + q^2 + q^3 + \cdots + q^n. \quad \text{(1)} $$
Multiplying Equation (1) by $q$:
$$ qS_n = q^2 + q^3 + q^4 + \cdots + q^{n+1}. \quad \text{(2)} $$
Subtracting (2) from (1), all interior terms cancel:
$$ S_n - qS_n = q - q^{n+1} \quad\Longrightarrow\quad S_n(1-q) = q - q^{n+1}. $$
Assuming $q \neq 1$, solving for $S_n$:
$$ S_n = \frac{q - q^{n+1}}{1-q} = \frac{q(1-q^n)}{1-q}. $$
Step 2 · Determine the sum $S$
The sum $S$ is the limit of $S_n$ as $n \to \infty$:
$$ S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{q - q^{n+1}}{1-q}. $$
Convergence hinges entirely on the behavior of $q^{n+1}$:
- Case $|q| < 1$ (convergence): $\lim_{n\to\infty} q^{n+1} = 0$, so $S = \dfrac{q-0}{1-q} = \dfrac{q}{1-q}$.
- Case $q = 1$ (divergence): $S_n = 1+1+\cdots+1 = n$, so $S_n \to \infty$.
- Case $q = -1$ (divergence): $S_n = -1+1-1+\cdots$ oscillates between $-1$ and $0$ and never settles.
- Case $|q| > 1$ (divergence): $|q^{n+1}| \to \infty$, so $S_n$ has no finite limit.
Hence $S$ exists (as a finite value) only when $|q| < 1$.
Summary Table
| Case | Behavior of $S_n$ as $n\to\infty$ | Sum $S$ |
|---|---|---|
| $|q| < 1$ | $q^{n+1}\to 0$ | $\dfrac{q}{1-q}$ |
| $q = 1$ | $S_n = n \to \infty$ | Diverges |
| $q = -1$ | Oscillates | Diverges |
| $|q| > 1$ | $|q^{n+1}|\to \infty$ | Diverges |
Additional Notes
- This series starts at $n=1$ rather than $n=0$, so its sum $\dfrac{q}{1-q}$ carries one fewer factor of $q$ than the more familiar $\sum_{n=0}^{\infty} q^n = \dfrac{1}{1-q}$; indeed $\dfrac{q}{1-q} = \dfrac{1}{1-q} - 1$.
- The formula $S_n = \dfrac{q(1-q^n)}{1-q}$ requires $q \neq 1$, since division by $1-q$ is undefined there; the $q=1$ case must be handled separately (and trivially, since every term equals $1$).
- At the boundary $|q|=1$, both sub-cases ($q=1$ and $q=-1$) fail to converge, confirming that convergence requires the strict inequality $|q|<1$.
Result
The partial sum is:
The sum of the series is: