Geometric Series

The central idea

A geometric series has the form $\sum_{n=0}^{\infty}ar^n$ or an index-shifted equivalent. It converges exactly when $|r|<1$, in which case its sum is $a/(1-r)$, where $a$ is the first included term. If $|r|\ge1$ and $a\ne0$, it diverges.

Definitions, hypotheses, and notation

For $r\ne1$, the finite identity $S_N=a(1-r^N)/(1-r)$ follows by subtracting $rS_N$ from $S_N$; when $r=1$, $S_N=Na$. The infinite formula is its limit, so it is valid only when $r^N\to0$. This derivation explains both the convergence condition and why substituting an inadmissible ratio into $a/(1-r)$ creates a meaningless answer.

Recurring decimals are geometric series: $0.272727\ldots=0.27+0.0027+\cdots$ has ratio $0.01$. Applications in finance and decay use the same structure, but units and starting time determine which payment or amount is the first term.

Conceptual meaning

Each term is a fixed multiple of the previous one. When $|r|<1$, the unadded tail shrinks geometrically. Negative $r$ alternates signs; the convergence condition depends on magnitude, while the sum formula retains the sign.

A dependable method and decision rule

1. Identify the first term actually included and common ratio.
2. Confirm the ratio is constant by dividing consecutive terms.
3. For a finite sum, use $S_N=a(1-r^N)/(1-r)$ when $r\ne1$, or $S_N=Na$ when $r=1$.
4. Check $|r|<1$ before using $S=a/(1-r)$ for an infinite sum.
5. For a tail beginning later, recompute its first term rather than reusing the original $a$.

Fully worked example

Graphical or geometric meaning

A unit interval can be filled by successively adding a fixed fraction of the remaining scale. Partial sums approach a horizontal level, and the gap to that level is itself a scaled geometric term.

Common mistakes and why they fail

Verification and reasonableness checks

• Multiply each term by $r$ and obtain the next.
• Verify the sum is plausible relative to the first term and sign pattern.
• Use the tail formula to check a partial-sum approximation.

Separate finite algebra from an infinite limit

For $r\ne1$, $S_N=a(1-r^N)/(1-r)$ is valid for every finite geometric sum, even when $|r|>1$. For $r=1$, the $N$ terms sum to $Na$. Only an infinite sum requires $|r|<1$, because then $r^N\to0$ and $S_N\to a/(1-r)$. Always identify the first term actually included; shifting the starting index changes $a$ even though $r$ is unchanged. Expand several terms from the proposed $a$ and $r$ to check the setup. For an infinite answer, also verify that the partial sums approach the claimed value rather than merely substituting into a formula whose convergence hypothesis may fail.

Practice

1. Sum $\sum_{n=0}^\infty(1/3)^n$.
2. Does $\sum_{n=1}^\infty2^n$ converge?
3. Sum $\sum_{n=2}^\infty(1/2)^n$.

Connections and next steps

• Infinite Series
• Power Series
• Telescoping Series