Based on the supplied transcript. Punctuation and paragraph breaks have been added, with repeated fillers and false starts reduced for readability. Mathematical expressions have been typeset where clear. The initial calculation error and its correction are retained.
All right, here we are in 11.2, picking a question. I think I’ll stick with 23. That sounds like a good one: 11.2.23.
Let me get a screenshot. I want to focus heavily on theorem-based questions.
I also want to focus on questions for which the author provides solutions or hints online—the pink ones.
If you’re serious about your education, you should take a different approach. I know some of you are serious and some aren’t. This is a breaking point.
I wish I could talk to professors because they don’t understand what I understand about how students cheat. I know how you cheat. I helped you cheat. Maybe my punishment is to help professors prevent it.
Anyway, we’re going through the book. Read the question backward, looking for keywords: convergent, divergent, and, most importantly, geometric series.
Let’s go back and find the geometric series.
On the exam, write something in shorthand such as “By the geometric series formula,” and then write the formula. You have to write the formula.
$$\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r},\qquad |r|<1.$$
This is where you seriously lose credit on exams. That’s 50% of the credit.
Now take $(-3)^{n-1}/4^n$. Let me make sure I copied it correctly before going forward.
For the formula to work, we need this expression to match that form. It’s straightforward. The series begins at $n=1$, and I need the exponent $n-1$.
I’ll add zero, write zero as $1-1$, remove the parentheses, and rearrange.
I hope you understand how beneficial copy and paste is. Now I group the exponent so I can use the algebraic operation.
The denominator becomes $4^{n-1}$ multiplied by $4$. I can bring the factor of $1/4$ out front.
Now we have the ratio $-3/4$ raised to the power $n-1$, and it looks the way it should.
$$\sum_{n=1}^{\infty}\frac{(-3)^{n-1}}{4^n}
=\frac14\sum_{n=1}^{\infty}\left(-\frac34\right)^{n-1}.$$
We see that $r=-3/4$, so $|r|=3/4<1$.
Thus, by the geometric series result—you have to cite your sources; we’re writing a paper here—the series converges.
I forgot about daylight saving time. That’s why I’m out of it. I thought, “It’s only seven. Why am I so tired?” Normally, it would be eight right now.
The series converges, so we find the sum.
The sum is then one over one minus $r$, with $r=-3/4$. That gives one over one plus three-fourths, which is four-sevenths.
Did I make a typo? Let’s check the back of the book. We did 11.3.23—no, 11.2.23. That’s why I couldn’t find the sum.
One-seventh. Oh, I forgot $a$. Typo. It’s $a/(1-r)$.
This is why you must be careful. I always tell you to write the formula, and I just missed the factor of one-fourth.
With that factor included, the sum is one-seventh.
$$\frac{a}{1-r}
=\frac{\frac14}{1-\left(-\frac34\right)}
=\frac17.$$
That’s geometric series. It’s straightforward arithmetic, but the question is about comprehending the theorem, citing it, and then doing the calculation.
The problem is that students get addicted to the puzzle—the arithmetic. That isn’t mathematics; it’s a consequence.
Setting it up, putting it together, finalizing it, and stating the correct justification—that’s mathematics. A calculator can do the arithmetic.
You’re addicted to being a calculator. Have you thought about that? “Mom, I’m a calculator. Aren’t you proud?”
That’s the difference between mathematics and arithmetic. Getting attached to the arithmetic side won’t help you. A calculator can do that.
This is why students think AI is effective. It’s a calculator. You aren’t in college to become a calculator. That’s a circus act, not impressive.
All right. That’s geometric series.
