Infinite Geometric Series from Partial Sums

From Finite Sums to an Infinite Sum

A finite geometric sum has a simple formula, but an infinite series is not evaluated by substituting infinity into that formula. Instead, its value is defined through the limit of its finite partial sums. For a geometric series, the finite formula makes this limit accessible: when the common ratio has absolute value less than one, the powers of the ratio shrink to zero.

We will focus on the geometric series whose first term is one:

1+r+r^2+r^3+\cdots.

Here $r$ is the common ratio: each term after the first is obtained by multiplying the preceding term by $r$. The condition $|r|<1$ is crucial. It ensures that the finite partial sums approach a limit, and it determines when the infinite-series formula derived below is valid.

Definition: For a real number $r$, the geometric series with first term one is $\sum_{k=0}^{\infty}r^k$. Its $N$th partial sum is $S_N=\sum_{k=0}^{N}r^k$. The series converges to $S$ if the sequence of partial sums has the finite limit $S$ as $N$ tends to infinity.

The index starts at zero, so $r^0=1$ is the first term. The partial sum $S_N$ includes the terms from $r^0$ through $r^N$, giving $N+1$ terms. Keeping this indexing clear matters: the exponent in the finite-sum formula will be $N+1$.

The Finite Partial-Sum Formula

For $r\ne 1$, the finite geometric-sum formula is

S_N=1+r+r^2+\cdots+r^N=\frac{1-r^{N+1}}{1-r}.

One way to see why this formula holds is to multiply the sum by $r$, then subtract. The two expressions have matching terms except for the first and last powers:

\begin{aligned} S_N&=1+r+r^2+\cdots+r^N,\\ rS_N&=r+r^2+r^3+\cdots+r^{N+1},\\ S_N-rS_N&=1-r^{N+1}. \end{aligned}

Since $S_N-rS_N=(1-r)S_N$, division by $1-r$ gives the formula. In the case considered here, $|r|<1$ implies $r\ne 1$, so its denominator is not zero.

Why the Powers Approach Zero

To take the limit of $S_N$, we need to know what happens to $r^{N+1}$. The key fact is that if $|r|<1$, then $r^n$ tends to zero as $n$ tends to infinity. Here is a direct justification.

Lemma: If $|r|<1$, then $\lim_{n\to\infty}r^n=0$.

If $r=0$, every positive power is zero, so the result holds immediately. Otherwise, let $q=|r|$, so $0<q<1$, and set $h=1/q-1$. Then $h>0$ and $1/q=1+h$. The binomial inequality $(1+h)^n\ge 1+nh$ follows by induction: it holds for $n=1$, and multiplying the inequality for $n$ by $1+h$ gives

(1+h)^{n+1}\ge(1+nh)(1+h) =1+(n+1)h+nh^2 \ge1+(n+1)h.

Taking reciprocals of positive quantities now gives

0\le q^n=\frac{1}{(1+h)^n}\le\frac{1}{1+nh}.

For any $\varepsilon>0$, choose a natural number $N$ large enough that $1/(1+Nh)<\varepsilon$. For every $n\ge N$, the bound above gives $0\le q^n<\varepsilon$. Thus $q^n$ tends to zero. Since $|r^n|=q^n$, it follows that $r^n$ also tends to zero.

Deriving the Infinite Sum

We can now take the limit of the finite partial sums. The argument uses both the finite formula and the condition $|r|<1$; without that condition, the same limit step is not justified.

Write the partial sum.
For each $N$, the sum of the first $N+1$ terms is $S_N=(1-r^{N+1})/(1-r)$, provided $r\ne1$.

Use the condition on the ratio.
When $|r|<1$, the lemma shows that $r^{N+1}$ tends to zero as $N$ tends to infinity.

Take the limit.
The denominator $1-r$ is fixed and nonzero, so the limit of the partial sums is $1/(1-r)$.

\begin{aligned} \lim_{N\to\infty}S_N &=\lim_{N\to\infty}\frac{1-r^{N+1}}{1-r}\\ &=\frac{1-0}{1-r}\\ &=\frac{1}{1-r}. \end{aligned}

Theorem: If $r$ is real and $|r|<1$, then the geometric series with first term one converges, and its sum is $$ \sum_{k=0}^{\infty}r^k=\frac{1}{1-r}. $$ This equality follows from the limit of the partial sums; the condition $|r|<1$ is required.

Worked Examples

Worked Example: The Ratio Is One Half

Find the value of $1+\frac12+\frac14+\frac18+\cdots$ using partial sums, and check the result.

Here $r=\frac12$, and $\left|\frac12\right|<1$, so the theorem applies. The $N$th partial sum is

S_N=\frac{1-(1/2)^{N+1}}{1-1/2}.

The denominator is $1-\frac12=\frac12$, so

S_N=2\left(1-\left(\frac12\right)^{N+1}\right).

As $N$ tends to infinity, $(1/2)^{N+1}$ tends to zero. Therefore

\lim_{N\to\infty}S_N=2(1-0)=2.

Check with a finite partial sum: for $N=3$, the terms are $1,\frac12,\frac14,\frac18$. Their sum is $\frac{8+4+2+1}{8}=\frac{15}{8}$. The formula also gives $S_3=2(1-\frac1{16})=\frac{15}{8}$. The limit value is consistent: $2-\frac{15}{8}=\frac18$, exactly the amount by which this partial sum falls short of $2$.

Worked Example: The Ratio Is Two Thirds

Find the sum of $1+\frac23+\frac49+\frac8{27}+\cdots$, and verify the formula using the first four terms.

Here $r=\frac23$, and $\left|\frac23\right|<1$. The partial sums satisfy

S_N=\frac{1-(2/3)^{N+1}}{1-2/3} =\frac{1-(2/3)^{N+1}}{1/3}.

The power $(2/3)^{N+1}$ tends to zero, so

\lim_{N\to\infty}S_N=\frac{1}{1/3}=3.

For the first four terms, $N=3$. Direct addition gives

1+\frac23+\frac49+\frac8{27} =\frac{27+18+12+8}{27} =\frac{65}{27}.

The partial-sum formula agrees:

S_3=\frac{1-(2/3)^4}{1/3} =\frac{1-16/81}{1/3} =\frac{65/81}{1/3} =\frac{65}{27}.

Finally, $3-\frac{65}{27}=\frac{81-65}{27}=\frac{16}{27}$, which checks the difference between the limiting value and this particular partial sum.

Worked Example: A Short Partial Sum and Its Limit

Consider $1+\frac14+\frac1{16}+\frac1{64}+\cdots$. Find the sum and check it against the partial sum through $r^2$.

The common ratio is $r=\frac14$, so $|r|<1$. The infinite sum is

\frac{1}{1-r} =\frac{1}{1-1/4} =\frac{1}{3/4} =\frac43.

The partial sum through $r^2$ is $S_2$, because its terms have exponents $0,1,2$:

S_2=1+\frac14+\frac1{16} =\frac{16+4+1}{16} =\frac{21}{16}.

Using the finite formula gives the same value:

S_2=\frac{1-(1/4)^3}{1-1/4} =\frac{1-1/64}{3/4} =\frac{63}{64}\cdot\frac43 =\frac{21}{16}.

The limiting value exceeds this partial sum by $\frac43-\frac{21}{16}=\frac{64-63}{48}=\frac1{48}$. This confirms the arithmetic for the selected finite sum while the limit calculation gives the value of the entire series.

Why the Condition Matters

The formula $\frac{1}{1-r}$ is not a rule that applies to every geometric series. It was obtained by taking a limit, and that step depended on $r^{N+1}$ tending to zero. For example, if $r=2$, the partial sums are

S_N=1+2+2^2+\cdots+2^N =\frac{1-2^{N+1}}{1-2} =2^{N+1}-1.

These partial sums grow without bound; they do not approach a finite value. Although substituting $r=2$ into $1/(1-r)$ produces $-1$, that number is not the sum of this series. The derivation of the infinite-sum formula does not apply because $|2|<1$ is false.

A common mistake is to quote the finite geometric formula, replace $r^{N+1}$ by zero, and conclude that the result holds for any ratio. The replacement is legitimate only when the powers actually tend to zero. A second indexing mistake is to use $r^N$ where the finite sum ends at $r^N$: the difference $S_N-rS_N$ contains $1-r^{N+1}$, not $1-r^N$.

Key takeaway: For the series $1+r+r^2+\cdots$, first check $|r|<1$. Then the finite partial sums have a limit, and that limit is $1/(1-r)$. The formula is a consequence of convergence, not a substitute for checking it.

Check Your Understanding

Use partial sums and the condition on the common ratio to answer these questions.

Write the $N$th partial sum of $1+\frac13+\frac19+\cdots$ using the finite geometric-sum formula.
What is the sum of $1+\frac13+\frac19+\cdots$, and which condition permits the limit step?
For $r=\frac12$, calculate $S_2$ both by adding the terms and by using the finite formula.
Explain why the formula $1/(1-r)$ cannot be used to assign a sum to $1+2+4+8+\cdots$.
In the expression for $S_N$, why does the numerator contain $r^{N+1}$ rather than $r^N$?