Mathematics¶
Geometric Series¶
🔷 Statement
For (\(|r| < 1\)):
\[
\sum_{i=0}^{\infty} r^i = \frac{1}{1 - r}
\]
🔹 Step 1: Start with a finite sum
Let:
\[
S_n = \sum_{i=0}^{n} r^i = 1 + r + r^2 + \cdots + r^n
\]
🔹 Step 2: Multiply by (\(r\))
\[
r S_n = r + r^2 + r^3 + \cdots + r^{n+1}
\]
🔹 Step 3: Subtract
\[
S_n - rS_n = (1 + r + r^2 + \cdots + r^n) - (r + r^2 + \cdots + r^{n+1})
\]
Everything cancels except:
\[
S_n - rS_n = 1 - r^{n+1}
\]
🔹 Step 4: Factor
\[
S_n(1 - r) = 1 - r^{n+1}
\]
So:
\[
S_n = \frac{1 - r^{n+1}}{1 - r}
\]
🔹 Step 5: Take the limit
If (\(|r| < 1\)), then (\(r^{n+1} \to 0\)) as (\(n \to \infty\)).
So:
\[
\lim_{n \to \infty} S_n = \frac{1}{1 - r}
\]
🔹 Final result
\[
\sum_{i=0}^{\infty} r^i = \frac{1}{1-r}, \quad |r|<1
\]