Geometric Series With Ratios Outside the Unit Interval

When a Geometric Series Grows

A geometric series is built by repeatedly multiplying each term by the same number. When the common ratio has absolute value less than one, successive terms shrink toward zero. At the boundary values \(1\) and \(-1\), the terms do not shrink. This tutorial considers the remaining case: a common ratio whose absolute value is greater than one.

In this case, each multiplication increases the magnitude of a nonzero term. The terms therefore fail a necessary condition for convergence of the series. The sign of the ratio affects whether the terms keep the same sign or alternate, but it does not prevent their magnitudes from growing.

Definition and the Main Conclusion

Definition: A geometric series with first coefficient \(a\) and common ratio \(r\) is the infinite series

\sum_{n=0}^{\infty} ar^n = a+ar+ar^2+ar^3+\cdots.

Its \(n\)th term, with this indexing, is \(u_n=ar^n\). The ratio between consecutive terms is \(r\), provided the terms being compared are nonzero.

The condition \(|r|>1\) means either \(r>1\) or \(r<-1\). The conclusion depends also on \(a\): if \(a\ne 0\), the terms grow in magnitude and the series diverges. If \(a=0\), every term is zero, so the series converges to zero regardless of \(r\).

Theorem: If \(a\ne 0\) and \(|r|>1\), then \(\sum_{n=0}^{\infty} ar^n\) diverges. If \(a=0\), the series is the zero series and converges.

Why Growing Terms Force Divergence

For a series to converge, its terms must approach zero. To see why, let \(S_N=u_0+u_1+\cdots+u_N\) be its \(N\)th partial sum. If the series converges, then the partial sums approach a finite limit. For \(N\ge 1\), the term \(u_N\) is the difference between two consecutive partial sums:

u_N=S_N-S_{N-1}.

As \(N\) grows, both partial sums on the right approach the same limit, so their difference approaches zero. Thus, if the terms do not approach zero, the partial sums cannot converge. This is the necessary-condition reasoning behind the nth-term test: a nonzero term limit, no term limit, or unbounded terms establishes divergence. A term limit of zero alone does not establish convergence.

Now suppose \(a\ne 0\) and \(|r|>1\). Set \(q=|r|\), so \(q=1+h\) for some \(h>0\). The binomial theorem gives, for every positive integer \(n\),

q^n=(1+h)^n\ge 1+nh.

Since \(1+nh\) grows without bound as \(n\) increases, so does \(q^n\). Therefore

|ar^n|=|a|\,|r|^n=|a|q^n\longrightarrow\infty,

because \(|a|>0\). In particular, the terms do not approach zero. The necessary condition for convergence fails, proving the theorem for \(a\ne 0\). If \(a=0\), instead, \(ar^n=0\) for every \(n\), which verifies the exception.

Worked Examples

Worked Example: A Positive Ratio Greater Than One

Determine whether the series \(\sum_{n=0}^{\infty} 2\cdot 3^n\) converges.

Here the first coefficient is \(a=2\) and the ratio is \(r=3\). We have \(a\ne0\) and \(|r|=3>1\). Its terms are

u_0=2\cdot3^0=2,\qquad u_1=2\cdot3^1=6,\qquad u_2=2\cdot3^2=18,\qquad u_3=2\cdot3^3=54.

Each term is obtained by multiplying the previous one by \(3\). More generally, \(|u_n|=2\cdot3^n\) grows without bound, so \(u_n\) does not tend to zero. The nth-term test therefore gives divergence.

Check: The listed terms have consecutive ratios \(6/2=3\), \(18/6=3\), and \(54/18=3\), as required. Their magnitudes increase, confirming that the terms do not approach zero.

Worked Example: A Negative Ratio with Growing Magnitudes

Determine whether the series \(\sum_{n=0}^{\infty} 4(-2)^n\) converges.

Here \(a=4\), \(r=-2\), and \(|r|=2>1\). The first several terms are

u_0=4,\qquad u_1=-8,\qquad u_2=16,\qquad u_3=-32,\qquad u_4=64.

The signs alternate, but the magnitudes are \(4,8,16,32,64,\ldots\), so the terms do not approach zero. In fact, their magnitudes are \(4\cdot2^n\), which grow without bound. The nth-term test proves that the series diverges.

One can also see the failure of convergence in the partial sums: \(S_0=4\), \(S_1=4-8=-4\), \(S_2=-4+16=12\), and \(S_3=12-32=-20\). These partial sums do not settle toward a finite value.

Check: The ratio of each term to the preceding term is \(-2\), and the partial sums satisfy \(S_3=-20\). Both calculations agree with the stated series and its alternating, growing behavior.

Worked Example: A Negative First Coefficient and a Fractional Ratio

Determine whether the series \(\sum_{n=0}^{\infty} -2\left(-\frac{3}{2}\right)^n\) converges.

The first coefficient is \(a=-2\), and the ratio is \(r=-\frac{3}{2}\). Since \(a\ne0\) and \(|r|=\frac{3}{2}>1\), the term magnitudes grow. Computing the first terms gives

u_0=-2,\qquad u_1=(-2)\left(-\frac32\right)=3,\qquad u_2=3\left(-\frac32\right)=-\frac92,\qquad u_3=-\frac92\left(-\frac32\right)=\frac{27}{4}.

The magnitudes are \(2,3,\frac92,\frac{27}{4}\). Each is \(\frac32\) times the previous magnitude. In general, \(|u_n|=2\left(\frac32\right)^n\), which grows without bound. Thus the terms do not tend to zero, and the series diverges by the nth-term test.

Check: The term ratios are \(3/(-2)=-\frac32\), \((-\frac92)/3=-\frac32\), and \((\frac{27}{4})/(-\frac92)=-\frac32\). The computed terms and their alternating signs match the given ratio.

Partial Sums as an Additional Check

The finite geometric-sum formula gives another way to inspect this behavior. For \(r\ne1\), the partial sum through \(u_N\) is

S_N=a+ar+\cdots+ar^N =\frac{a(1-r^{N+1})}{1-r}.

When \(a\ne0\) and \(|r|>1\), the power \(r^{N+1}\) grows in magnitude. Consequently the partial sums cannot approach a finite limit. For example, in the series \(4+4(-2)+4(-2)^2+\cdots\), the formula gives

S_3=\frac{4(1-(-2)^4)}{1-(-2)} =\frac{4(1-16)}{3} =-20,

which agrees with the direct calculation \(4-8+16-32=-20\). This formula is a useful check, but the decisive convergence test is already visible in the individual terms: they do not approach zero.

A Common Mistake: Treating Alternation as Cancellation

A negative ratio does make consecutive terms alternate in sign, but alternation alone does not imply convergence. For instance, a mistaken argument might say that the terms \(4,-8,16,-32,\ldots\) cancel in pairs, and therefore the series converges. The first two terms do sum to \(-4\), but the next two sum to \(-16\):

4+(-8)=-4,\qquad 16+(-32)=-16.

The pair sums do not approach zero; their magnitudes grow. More fundamentally, the original terms have magnitudes \(4,8,16,32,\ldots\), so the terms fail the necessary condition for convergence. Pairing terms cannot repair that failure.

Another condition worth checking is \(a\ne0\). The ratio by itself does not guarantee divergence if the first coefficient is zero. For example, if \(a=0\) and \(r=2\), then \(ar^n=0\) for every \(n\), and the series sums to zero. Always identify both the coefficient and the ratio before applying the theorem.

What the Ratio Tells You

For a nonzero first coefficient, a ratio with magnitude greater than one forces successive term magnitudes to increase by a fixed factor. A positive ratio produces terms with a consistent sign; a negative ratio produces alternating signs. In either case, the magnitudes grow, so the terms cannot approach zero and the series diverges.

This is a direct use of the nth-term test: it can establish divergence when the terms fail to tend to zero. It cannot establish convergence merely because the terms do tend to zero. The essential check here is therefore not just the sign pattern but the behavior of \(|ar^n|\).

Check Your Understanding

For each question, use the first coefficient, the common ratio, and the behavior of the terms.

Does \(\sum_{n=0}^{\infty} 5\left(\frac43\right)^n\) converge or diverge? State the term behavior that supports your answer.
Does \(\sum_{n=0}^{\infty} 3(-4)^n\) converge or diverge? Explain why alternating signs do not change the conclusion.
Why must the terms of a convergent series approach zero?
What is the outcome of \(\sum_{n=0}^{\infty}0\cdot 7^n\), even though the ratio has magnitude greater than one?
For \(a\ne0\), what does \(|r|>1\) imply about \(|ar^n|\) as \(n\) increases?