Infinitely many pieces, one finite square.

Keep adding forever and you might expect the total to grow forever. For a geometric series with the right ratio, it doesn't: the pieces fit inside a square with room to spare, and the total is exact.

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Shade half the square, then half of what is left, then half of that. Each piece is half the one before, so their areas make a geometric series with first term $a = \tfrac{1}{2}$ and common ratio $r = \tfrac{1}{2}$:

The running totals are $\tfrac{1}{2}, \tfrac{3}{4}, \tfrac{7}{8}, \tfrac{15}{16}, \dots$ Each new piece fills half of the gap to $1$, so after $n$ pieces the gap is exactly $(\tfrac{1}{2})^n$: the size of the last piece.

Look closer and the halving never stops. The gap can be made as small as you like, so the infinite sum is exactly $1$. That value is the sum to infinity:

The limit comes from $r^n \to 0$, which needs $|r| < 1$. With $r = 2$ the pieces grow instead, $1 + 2 + 4 + \dots$, and the totals never settle: that series has no sum to infinity.

Multiply by r, subtract, and almost everything cancels.

The $n$th term, recurring decimals as fractions, modelling growth and decay, and twelve graded exam questions with full worked solutions are waiting.

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