Every transform this course builds is a summation. That covers the
z-transform, the DFT, and chapter 2's central convolution formula. Most of
them either are a geometric series outright, or reduce to one at a key
step. This section derives the one fact used constantly from chapter 2
onward. It does so in the same summation shorthand this course leans on.
Chapter 0 of fourier-analysis is where this was first met. The version
stated there runs over the real numbers rather than over a discrete
index.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Probability
A geometric series has first term a=5 and common ratio r=0.25. What does the series converge to? Round to 2 decimal places.
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Why
a=5, r=0.25: sum = a/(1-r) = 5/0.75 ≈ 6.67.
Correct answer
Also accepted:
Why
Plug a and r directly into a/(1-r).
From Probability
A ball is dropped and each bounce reaches 0.6 of the height of the previous bounce, starting at 10 feet. What is the total of all the (infinitely many) bounce heights after the first drop, \sum_{k=1}^{\infty} 10(0.6)^k? Round to 1 decimal place.
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Why
This is a/(1-r) with a = 10(0.6) = 6 and r = 0.6: 6/(1-0.6) = 6/0.4 = 15.
Correct answer
Also accepted:
Why
The sum starts at k=1, not k=0, so the first term (a) is 10(0.6), not 10.
From Fourier Analysis
Evaluate \displaystyle\sum_{n=1}^{\infty}\left(\frac14\right)^n.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Geometric with a=\frac14, r=\frac14: \frac{a}{1-r} = \frac{1/4}{3/4} = \frac13.
Correct answer
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Why
Pick a number and call it r. Start with 1. Multiply by r to get the next term, then by r again for the term after that. You get 1 + r + r^2 + r^3 and onward forever. Adding all of that up is what a geometric series means. The small raised numbers only count how many times you multiplied.
Does an endless sum like that add up to anything at all? It depends only on how big r is, ignoring its sign. That size is written |r|, where the two upright bars mean "distance from zero."
Take r=0.5, so |r| is less than 1. Each new term is half the term before it. The running total goes 1, then 1.5, then 1.75, then 1.875. It creeps toward 2 and never quite gets there. That is what "settles down to a fixed number" means. The sum has an answer.
Now take |r| \ge 1 instead. The terms never shrink. The running total either grows past any size you name, or it refuses to settle anywhere at all. In that case the endless sum has no answer.
The subtraction trick behind the finite formula is nothing fancy. Write the series out, then write a copy of it shifted over by one slot. Almost every term in the first list is matched in the second, so subtracting wipes those out. Only a first term and a last term survive. That is why so short a formula can stand in for so long a sum.
Both closed forms above say the same useful thing. Do not add up a pile of terms by hand. Plug in r and read off the exact total in one step.
A geometric series \sum_{n=0}^{\infty} r^n converges exactly when |r|<1, and it then has the closed form \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}, \qquad |r|<1. To see why, first find the finite sum S_N=\sum_{n=0}^{N-1}r^n, which runs over n=0 to N-1. Multiplying by r shifts every exponent up by one, giving rS_N=\sum_{n=1}^{N}r^n. Subtracting then cancels every term except the first and the last: (1-r)S_N = 1-r^N. So for any r\ne1, \sum_{n=0}^{N-1} r^n = \frac{1-r^N}{1-r} holds with no convergence condition attached, since it is a finite sum. Now let N\to\infty. When |r|<1, the power r^N\to0, and the finite sum settles at \frac{1}{1-r}, the infinite formula above. When |r|\ge1, that power never shrinks and the sum has no limit. This course's summation index almost always plays the role of a discrete-time index n. The ratio r is almost always a complex exponential, either r = e^{-j\omega} or r=z^{-1}. Chapters 3–4 substitute directly into the two closed forms above.
Watching a geometric sum settle down
This is the running total of 1 + r + r^2 + \cdots for a fixed r inside (-1, 1). Confirm it climbs toward the closed-form value 1/(1-r) rather than growing without bound.
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What divergence looks like running
This is the same running total for r=1.2, outside (-1, 1). Confirm it never settles -- it grows without bound instead, exactly as the |r|\ge1 case above predicts.
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Worked example
A finite geometric sum, the z-transform's own shape
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1
Evaluate \sum_{n=0}^{4} (0.5)^n using the finite-sum closed form. Apply \sum_{n=0}^{N-1} r^n = \dfrac{1-r^N}{1-r} with N=5, r=0.5.
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2\frac{1-(0.5)^5}{1-0.5} = \frac{1-0.03125}{0.5} = \frac{0.96875}{0.5}
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3\boxed{\sum_{n=0}^{4}(0.5)^n = 1.9375}
Chapter 4 computes exactly this shape of sum, with r = z^{-1} instead of a plain number, to find the z-transform of a finite-length sequence.
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Worked example
A complex-exponential ratio, previewing chapter 3
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1
Does \sum_{n=0}^{\infty} \left(\tfrac12 e^{j\pi/4}\right)^n converge? Convergence of \sum r^n depends only on |r|, not on r's angle. So compute the modulus.
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2
\left|\tfrac12 e^{j\pi/4}\right| = \tfrac12 \cdot \left|e^{j\pi/4}\right| = \tfrac12 \cdot 1 = \tfrac12, since every point e^{j\theta} lies on the unit circle.
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3\boxed{\text{Yes -- }|r|=\tfrac12<1\text{, so the series converges}}
The angle \pi/4 affects where the sum lands as a complex number. It never affects whether the series converges. Only the modulus of r governs that, exactly as it does for a real-valued ratio.
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Where this goes next
The z-transform's region of convergence, chapter 4
Chapter 4's z-transform is, term by term, exactly this kind of series. It is written X(z) = \sum_n x[n] z^{-n}. Its region of convergence is precisely the set of z for which the underlying geometric-series size condition |r|<1 holds. Every ROC computed in that chapter traces back to the one test reviewed here.
Check your understanding (counts toward your progress)
For which values of r does \sum_{n=0}^{\infty} r^n converge?
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That's exactly the divergence condition — the terms grow rather than shrink.
The convergence test depends only on |r|, real or complex.
An infinite geometric series diverges whenever |r|\ge 1.
Why
An infinite geometric series \sum r^n converges exactly when |r|<1, real or complex.
Evaluate the finite sum \sum_{n=0}^{2} (0.5)^n.
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Why
1 + 0.5 + 0.25 = 1.75, matching \dfrac{1-(0.5)^3}{1-0.5} = \dfrac{0.875}{0.5} = 1.75.
Correct answer
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Why
Either add the three terms directly, or use the finite-sum closed form with N=3, r=0.5.
Evaluate \sum_{n=0}^{\infty} (0.25)^n using the closed form \frac{1}{1-r}.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\dfrac{1}{1-0.25} = \dfrac{1}{0.75} = \dfrac{4}{3}.
Correct answer
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Why
Practice: Finite and infinite geometric sums (2 problems · doesn’t affect your progress)
Compute the finite sum \sum_{n=0}^{3} (0.5)^n.
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Why
1+0.5+0.25+0.125 = 1.875, matching \dfrac{1-(0.5)^4}{1-0.5}=\dfrac{0.9375}{0.5}=1.875.
Correct answer
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Why
Does \sum_{n=0}^{\infty} \left(\tfrac34 e^{j\pi/6}\right)^n converge?
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The modulus test applies to a complex ratio exactly as it does to a real one — only |r| matters, not whether r itself is real or complex.
The modulus of the ratio is computable directly — \left|\tfrac34 e^{j\pi/6}\right| = \tfrac34\cdot 1 = \tfrac34.
Why
Convergence depends only on the modulus of the ratio, \left|\tfrac34 e^{j\pi/6}\right| = \tfrac34 < 1, so the series converges regardless of the angle \pi/6.
Yes -- the modulus of the ratio is \tfrac34 < 1