The limiting sum of a geometric series
When an infinite geometric series has a limiting sum, the condition −1 < r < 1, the formula S = a/(1 − r), and problems including finding the values of x for which a limiting sum exists.
Adding infinitely many terms
Walk halfway to a wall, then half the remaining distance, then half again. The distances $\tfrac12 + \tfrac14 + \tfrac18 + \cdots$ never stop, but their total gets as close as you like to $1$. We say the series has limiting sum $1$.
Each step is one more partial sum S₁, S₂, S₃, … The dashed line is the limiting sum. With a negative ratio, the sums jump above and below it.
- Make $r$ negative. How do the steps change?
- Move $r$ close to $0.8$. Why does it take longer to get close?
Finding limiting sums
Find the limiting sum of $18 + 6 + 2 + \cdots$
Find the limiting sum of $8 - 4 + 2 - \cdots$
For which $x$ does $1 + (x - 1) + (x - 1)^2 + \cdots$ have a limiting sum? Find it when $x = 1.5$.
A GP has first term $4$ and limiting sum $12$. Find $r$.
Find the limiting sum of $18 + 6 + 2 + \cdots$
$\dfrac{18}{\frac23}$.
Find the limiting sum of $8 - 4 + 2 - \cdots$
$\dfrac{8}{\frac32}$.
A GP has first term $4$ and limiting sum $12$. Find $r$.
$1 - r = \dfrac13$.
For which $x$ does $1 + (x - 1) + (x - 1)^2 + \cdots$ have a limiting sum?
$-1 < x - 1 < 1$.
Which series has a limiting sum?
Its ratio is $-0.8$.
Find the limiting sum of $0.9 + 0.09 + 0.009 + \cdots$
$\dfrac{0.9}{1 - 0.1}$.
Find the limiting sum of $5 + \dfrac52 + \dfrac54 + \cdots$
$\dfrac{5}{\frac12}$.
A GP has ratio $\dfrac14$ and limiting sum $8$. Find its first term.
$a = 8 \times \dfrac34$.
Find the limiting sum of $1 + (x - 1) + (x - 1)^2 + \cdots$ when $x = 1.5$.
$\dfrac{1}{2 - 1.5}$.
New numbers every time. Keep going until you get five in a row.
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