Infinite Geometric Series

Adding infinitely many terms sounds like it must blow up to infinity, but it does not always work that way. When the ratio is smaller than 1, each new piece is a fraction of the one before, so no matter how many you add the total never breaks through a certain ceiling and instead converges to it. That ceiling is exactly a/(1−r), the first term divided by one minus the ratio. Drag the term-count n and watch the partial-sum bars keep climbing yet never cross the red limit line.

Adding Forever, Yet It Ends?

Terms added n4

👀 See It

①Adding more terms grows the bar

②But if the ratio is below 1, each added amount shrinks

③So even adding forever, it never passes a certain value (the red line) and converges to it

Start From the Partial Sum

Partial sum of a geometric sequence

S_n = a(1 - rⁿ)1 - r (r ≠ 1)

Sum of the first n terms with first term a and ratio r

🧱 First, the Sum of n Terms

①An infinite series is defined as the limit of partial sums Sₙ

②Write the geometric partial-sum formula first

③Then send n→∞ — this order is the key

Take the Limit and the Formula Appears

Sum of an infinite geometric series

S = a1 - r (|r| < 1)

With first term a and ratio r, it converges only if |r|<1, with sum a/(1−r)

Convergence condition

converges ⇔ a = 0 or |r| < 1

The first term is 0, or the ratio has absolute value below 1

💨 rⁿ Vanishes

①If |r|<1, then rⁿ→0 as n→∞

②The rⁿ term disappears from the partial sum

③What remains is the infinite-geometric-series formula

Compute It Directly

Example 1

Find the sum of the infinite geometric series with first term 3 and ratio 1/3.

Check the ratio — since |1/3|<1, it converges.

Substitute a=3, r=1/3 into S = a/(1−r).

S = 31 - 1/3 = 32/3

▸ S = 9/2

Checking |r|<1 convergence before applying the formula prevents mistakes.

Wrap-up

Key result

|r|<1 ⇒ ∑_n=1^∞ a r^n-1 = a1 - r

An infinite geometric series with first term a, ratio r converges to a/(1−r) when |r|<1

2023 KICE mock exam Math (Calculus) type, adapted

Find ∑_n=1^∞ 2ⁿ3ⁿ.

①1

②3/2

③2

④5/2

⑤Diverges

▸ ③ 2

Reading the general term as (2/3)ⁿ, it is a geometric series with first term 2/3 and ratio 2/3.

Since |2/3|<1, S = (2/3)/(1−2/3).

S = 2/31/3 = 2

🎯 Exam Points

①Always check the |r|<1 convergence condition first

②The sum is S=a/(1−r); pin down the first term and ratio exactly

③Mind the first-term difference between ∑arⁿ and ∑arⁿ⁻¹

④Repeating decimals and figure sums reduce to this formula too

⑤If it does not converge, the sum does not exist

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Series

Applications of Geometric Series

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