Proof of the Geometric Series Formula

From Finite Sums to an Infinite Sum

The previous tutorial established two facts that now fit together: the Finite Geometric-Sum Identity gives an exact expression for each partial sum, and the Convergence Criterion for a Geometric Series tells us that the series converges when $|r|<1$. This tutorial uses those facts to identify the limit of the partial sums. It also turns the same calculation into an exact formula for the error after truncation.

For a geometric series $\sum_{n=0}^{\infty} ar^n$, write $S_N=\sum_{n=0}^{N} ar^n$. The index $N$ marks the last term included, so $S_N$ contains $N+1$ terms. When $|r|<1$, we have $r\ne1$, and the finite identity gives $S_N=a(1-r^{N+1})/(1-r)$. The only remaining step is to determine the limit of this expression as $N$ increases.

Theorem (Sum of a Convergent Geometric Series): If $|r|<1$, then the geometric series $\sum_{n=0}^{\infty} ar^n$ converges and $$ \sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}. $$ This includes $a=0$ and $r=0$.

Proof. Fix $a\in\mathbb{R}$ and suppose $|r|<1$. By the Convergence Criterion for a Geometric Series, the series converges. Denote its sum by $S$, so $S_N\to S$. For every $N\geq0$, the Finite Geometric-Sum Identity gives

S_N =\sum_{n=0}^{N} ar^n =a\frac{1-r^{N+1}}{1-r} =\frac{a}{1-r}-\frac{a r^{N+1}}{1-r}.

The denominator $1-r$ is nonzero because $|r|<1$ rules out $r=1$. Also, $r^{N+1}\to0$ when $|r|<1$, as used in the convergence proof in the previous tutorial. Therefore the second term on the right tends to zero, and the first term is constant. Taking the limit of the displayed equality yields

S=\lim_{N\to\infty}S_N=\frac{a}{1-r}.

If $a=0$, every partial sum is zero, and the formula gives $0/(1-r)=0$, as required. If $r=0$, the series consists of $a$ followed by zeros, and the formula gives $a/(1-0)=a$. Thus both cases are included. $\square$

The formula depends on taking a limit; it is not an identity for an infinite sum in the same way that the finite-sum formula is an identity for each $S_N$. The condition $|r|<1$ ensures both that the series converges and that the power in the partial-sum expression disappears in the limit.

Worked Example: A Positive Ratio

Consider $\sum_{n=0}^{\infty}6(2/5)^n$. Since $|2/5|<1$, the theorem applies, and the sum is

\sum_{n=0}^{\infty}6\left(\frac25\right)^n =\frac{6}{1-2/5} =\frac{6}{3/5} =10.

As a finite check, the partial sum through $n=3$ is

S_3=6+ \frac{12}{5}+\frac{24}{25}+\frac{48}{125} =\frac{750+300+120+48}{125} =\frac{1218}{125}.

This is $9.744$, which is below $10$. The terms omitted after $n=3$ have total $10-1218/125=32/125$. The next result gives that error directly, without first subtracting the partial sum from the full sum.

The Exact Remainder After Truncation

Once a series has a known sum, its remainder after $S_N$ is the difference between the sum and that partial sum. For a geometric series, the omitted terms form another geometric series with first term $ar^{N+1}$ and ratio $r$. The remainder formula below follows directly from the sum theorem; it also makes the effect of the first omitted term explicit.

Corollary (Exact Geometric Remainder): If $|r|<1$ and $S=\sum_{n=0}^{\infty} ar^n$, then for every integer $N\geq0$, $$ R_N:=S-S_N=\frac{a r^{N+1}}{1-r}. $$ In particular, $$ |R_N|=\frac{|a|\,|r|^{N+1}}{|1-r|}. $$

Proof. Use the sum formula just proved and the Finite Geometric-Sum Identity for $S_N$:

R_N =S-S_N =\frac{a}{1-r}-a\frac{1-r^{N+1}}{1-r} =\frac{a-a+a r^{N+1}}{1-r} =\frac{a r^{N+1}}{1-r}.

Taking absolute values gives the stated expression for $|R_N|$, since $|1-r|$ is positive and nonzero. $\square$

This is an exact error, not merely an upper estimate. For example, if a desired error tolerance is $\varepsilon>0$, it is enough to choose $N$ so that \[ \frac{|a|\,|r|^{N+1}}{|1-r|}<\varepsilon. \] The index $N+1$ appears because it is the first omitted exponent when $S_N$ includes terms from $0$ through $N$. Confusing the last included term with the first omitted term is a common source of off-by-one errors.

The sign of $R_N$ also describes the position of the partial sum relative to the full sum. If $a>0$ and $0<r<1$, then $R_N>0$, so $S_N<S$. If $a>0$ and $-1<r<0$, the sign alternates with $N+1$; a partial sum may lie above or below the total. The absolute-value formula gives the size of the error in either case.

Worked Example: Alternating Terms and a Signed Remainder

Consider $\sum_{n=0}^{\infty}3(-1/4)^n$. Its ratio has absolute value $1/4<1$, so its sum is

S=\frac{3}{1-(-1/4)} =\frac{3}{5/4} =\frac{12}{5}.

The partial sum through $n=2$ is

S_2=3-\frac34+\frac{3}{16} =\frac{48-12+3}{16} =\frac{39}{16}.

The exact remainder is

R_2 =\frac{3(-1/4)^3}{1+1/4} =\frac{-3/64}{5/4} =-\frac{3}{80}.

Indeed, $S_2+R_2=39/16-3/80=195/80-3/80=192/80=12/5$. The negative remainder shows that $S_2$ is greater than the sum. Its absolute size, $3/80$, agrees with the difference $39/16-12/5=195/80-192/80=3/80$.

Using the Remainder to Choose a Partial Sum

The remainder formula can be used in the forward direction: first choose an error tolerance, then find a partial sum whose error is smaller than that tolerance. Because $|r|^{N+1}\to0$ for $|r|<1$, such an $N$ always exists. The formula indicates why ratios close to $1$ in absolute value can require many terms: their powers decrease more slowly.

Worked Example: Choosing a Truncation for a Prescribed Error

Consider $\sum_{n=0}^{\infty}(9/10)^n$. Its sum is $1/(1-9/10)=10$. We seek a partial sum with error less than $1/2$. For the partial sum through $n=29$, the exact remainder formula gives

|R_{29}| =\frac{(9/10)^{30}}{1/10} =10\left(\frac{9}{10}\right)^{30} \approx 0.4239 <\frac12.

Thus the first 30 terms, from exponent $0$ through exponent $29$, approximate the sum $10$ to within $1/2$. This claim concerns the actual error, not a guessed decimal approximation: the exact error is $10(9/10)^{30}$, and its displayed decimal value verifies the required strict inequality.

The formula also clarifies why it is not enough to inspect only the next term and assume that it equals the remaining error. The first omitted term is $ar^{N+1}$, while the remainder includes that term and every later one. In fact, when $|r|<1$, their exact combined value is $ar^{N+1}/(1-r)$. If $r$ is positive and close to $1$, the factor $1/(1-r)$ can be large, so the total omitted tail can be much larger than its first term.

Worked Example: A Negative Initial Term

Take $\sum_{n=0}^{\infty}-5(1/2)^n$. Since $|1/2|<1$, its sum is

S=\frac{-5}{1-1/2}=-10.

The partial sum through $n=3$ is

S_3=-5-\frac52-\frac54-\frac58 =-\frac{40+20+10+5}{8} =-\frac{75}{8}.

The remainder is

R_3=\frac{-5(1/2)^4}{1-1/2} =\frac{-5/16}{1/2} =-\frac58.

Adding the remainder checks the result: $-75/8-5/8=-80/8=-10$. Here the omitted terms are all negative, so the partial sum is greater than the full sum, even though both are negative.

What the Formula Does—and Does Not—Say

The sum formula is stated for $|r|<1$, the precise convergence range when $a\ne0$. At $r=1$, the denominator $1-r$ is zero, and the series with $a\ne0$ does not converge. At $r=-1$, the denominator is nonzero, but the series with $a\ne0$ still does not converge. A defined expression alone does not guarantee that it represents a series sum; the convergence hypothesis is essential.

There is one harmless exceptional case outside that range: when $a=0$, all terms vanish, so the series converges to zero for every real $r$. The expression $a/(1-r)$ still cannot be used at $r=1$, because it would have zero denominator. For $a=0$ and $r\ne1$, the expression evaluates to zero, consistently with the sum.

The exact remainder is useful both theoretically and computationally. It proves that partial sums approach the stated value, shows how many terms are needed for a given accuracy, and records whether an approximation lies above or below the sum. In the next tutorial, the same attention to finite partial sums will be applied to series whose terms cancel in a different pattern.

Takeaway: For $|r|<1$, the finite partial sums converge to $a/(1-r)$. After $S_N$, the exact remainder is $ar^{N+1}/(1-r)$, so its absolute size gives a direct and precise measure of truncation error.

Check Your Understanding

Use the sum theorem and the exact remainder formula to answer the following questions.

Why does the finite partial-sum expression have the exponent $N+1$ when $S_N$ ends with the term $ar^N$?
Find the sum of $\sum_{n=0}^{\infty}4(1/3)^n$.
For $\sum_{n=0}^{\infty}2(-1/3)^n$, find the exact remainder after $S_2$ and state whether $S_2$ is above or below the sum.
Why must $|r|<1$ be stated when using $a/(1-r)$ as the sum of a geometric series?
If $|r|$ is close to $1$, what does the remainder formula suggest about the number of terms needed for a small error?