Calculus II ep20: Increasing/decreasing (Oct 28, 2026)

Overview

The lesson develops sequence limits and monotonicity, showing that absolute values handle alternating sequences such as (-1)^n/n and classifying the limits of r^n by the value of r. It then defines increasing and decreasing sequences and demonstrates proofs by comparing a_n with a_(n+1), including 3n/(n+1) and n/(n²+1).

Key takeaways

For any sequence, if the absolute-value sequence tends to zero, then the original sequence tends to zero; this proves (-1)^n/n→0 because its absolute value is 1/n.
The power sequence r^n converges to zero when |r|<1, equals 1 for all n when r=1, diverges to positive infinity when r>1, and diverges by oscillation when r≤-1.
A sequence is strictly increasing only if a_n<a_(n+1) for every index n, and strictly decreasing only if a_n>a_(n+1) for every n; checking a few initial terms is not a proof.
To prove 3n/(n+1) is increasing, compare it with 3(n+1)/(n+2); positive denominators allow cross-multiplication, and simplification gives the always-true statement 0<3.
To prove n/(n²+1) is decreasing, compare consecutive terms and simplify to n²+n>1, which holds for every positive integer n because its minimum value is 2.
When algebra simplifies a monotonicity proof to an obvious inequality, include words explaining that the simplified statement is equivalent to the original consecutive-term comparison.

Chapters

0:00 Alternating Sequence (-1)^n/n and the Absolute-Value Limit Trick

The sequence (-1)^n/n begins -1, 1/2, -1/3, 1/4, so its signs alternate while its terms approach zero.
Because (-1)^x is not real-valued for general real x, standard function-limit substitutions do not directly handle this sequence.
To prove a sequence tends to zero, show its absolute values tend to zero; here |(-1)^n/n| simplifies to 1/n, whose limit is zero.

8:48 Initial Examples of the Power Sequence r^n

For r=0, every term of r^n is zero; for r=1, every term is one.
When r=1/2, the terms 1/2, 1/4, 1/8, 1/16 approach zero as powers shrink.
When r=2, the terms 2, 4, 8, 16 grow without bound.

13:01 Positive r Cases: Convergence and Divergence of r^n

For r>1, successive powers increase without bound, so r^n diverges to positive infinity.
For r=1, the constant sequence r^n converges to 1.
For 0≤r<1, powers of r converge to zero, including the special case r=0.

15:11 Negative r Cases and the Complete Classification of r^n

For r=-1/2, the sequence -1/2, 1/4, -1/8, 1/16 alternates signs while approaching zero.
For r=-1, terms alternate between -1 and 1, so no single limit exists; for r=-2, alternating terms also grow in magnitude.
The complete rule is: |r|<1 gives limit 0, r=1 gives limit 1, r>1 diverges to infinity, and r≤-1 diverges without a limit.

22:14 Definitions of Increasing and Decreasing Sequences

A sequence is increasing when a_n<a_(n+1) for every n; the comparison must hold for all consecutive terms.
A sequence is decreasing when a_n>a_(n+1) for every n.
Examples include 1, 3, 5, 7 as increasing and 5, 0, -5, -10 as decreasing.

27:46 Monotonicity and Sequences That Are Neither Increasing nor Decreasing

Monotonic or monotone means a sequence is increasing or decreasing.
The sequence (-1)^n is not monotonic: it decreases from 1 to -1, then increases from -1 to 1.
A sequence can fail both definitions, so decreasing does not simply mean “not increasing.”

30:36 Testing 3n/(n+1) by Examining Its First Terms

For a_n=3n/(n+1), the first three terms are 3/2, 2, and 9/4.
Those values suggest the sequence is increasing, but checking a few terms does not prove the inequality holds for every n.
A proof must establish a_n<a_(n+1) using the sequence definition.

33:16 Proving 3n/(n+1) Is Increasing

Substitute n+1 into the formula to compare 3n/(n+1) with 3(n+1)/(n+2).
Since n+1 and n+2 are positive for sequence indices n≥1, cross-multiplication preserves the inequality direction.
Expanding and canceling equivalent terms reduces the comparison to 0<3, which is true for every n; therefore the sequence is increasing.

39:45 Practice Proof: Show n/(n²+1) Is Decreasing

The exercise asks for a proof that a_n=n/(n²+1) is decreasing, so the required comparison is a_n>a_(n+1).
The proof setup replaces n with n+1 throughout the formula, producing (n+1)/((n+1)²+1).
The guidance emphasizes writing a few explanatory words to identify the comparison and signal the algebraic simplification.

46:01 Completing the Decreasing Proof and Explaining Proof Logic

Cross-multiplying the positive denominators and expanding the comparison for n/(n²+1) yields n²+n>1.
For every positive integer n≥1, n²+n is at least 2, so a_n>a_(n+1) and the sequence is decreasing.
The algebra may simplify a desired inequality to an obviously true statement; explanatory words clarify that the proof establishes the original claim, not merely the final inequality.

Transcript excerpt

0:00 We been talking about sequences. Uh so, a little reminder, we do have homework for tonight as usual, and the quiz tomorrow will feature three questions. Uh they shall be partial fractions, improper integrals, and integration gumbo. So, be ready for the gumbo. It's going to be great. Um let's continue about sequences if everybody's ready for that. So, we were talking at the very end last time about um some special sort of ways that you got to handle -1 to the n as a sequence. This alternates the signs in a sequence. And this is a very common thing that

0:50 we're going to see often in the sequences that we talk about. Like, for example, -1 to the n * 1 / n. That means the sequence usually we call this something like an. That means the sequence looks like 1 / n, but the signs flip back and forth. And since it's a -1 to the n, it means term number one has a sign of -1 to the one, which is -1. So, this the terms would be -1 / 1, and then positive 1/2, and then -1/3, and then positive 1/4, and so on, right? Uh the -1 to the n alternates the signs, and then you just get whatever you get from the rest of it is in this case 1 / n. Um this could make it a little tricky to decide what the limit is because we usually think of this thing when we're

1:39 trying to think of the limit. We think of this thing as a function, as if it was a function of x, and then just take the limit of that using our our old tricks. But, uh that doesn't really work for something like this, -1 to the power

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