Decreasing Sequences

What "decreasing" means

A sequence that is always increasing or always decreasing is called monotonic. Here is what a decreasing sequence looks like when you plot its terms: every point sits lower than the one before.

The quickest way to get a feel for a sequence is to list its first few terms. For $a_n = 5 - 2n$ you get $3, 1, -1, -3, \ldots$, which looks decreasing. For $b_n = n^2 - 4$ you get $-3, 0, 5, 12, \ldots$, which is increasing. For $d_n = (-1)^n n$ you get $-1, 2, -3, 4, \ldots$, which jumps up and down, so it is neither increasing nor decreasing.

Listing terms gives you evidence, never a proof. A pattern in the first four terms can break at the fifth.

The difference test

To prove that $a_{n+1} < a_n$ for every $n$, look at the sign of the difference.

The recipe has three steps:

1. Write $a_{n+1}$ by replacing every $n$ with $(n+1)$.

2. Subtract $a_n$ and simplify.

3. Decide the sign of the result for all $n \ge 1$.

This pattern is always true for sequences $a_n = a_1 + (n-1)d$ with constant difference $d$: the difference $a_{n+1} - a_n$ equals $d$. So an arithmetic sequence is strictly decreasing exactly when $d < 0$.

When a sequence is given by a recurrence $a_{n+1} = a_n + (\text{something})$, you can read the difference directly. For $a_1 = 8$ and $a_{n+1} = a_n - \dfrac{1}{2^n}$, we get $a_{n+1} - a_n = -\dfrac{1}{2^n} < 0$ for every $n$. No formula for $a_n$ is needed.

Rational sequences

When $a_n$ is a fraction, the difference $a_{n+1} - a_n$ is a subtraction of fractions. Use a common denominator, and remember that the denominators are positive, so the sign of the whole thing is the sign of the numerator.

Notice that the $n^2$ and $n$ terms canceled. That always happens with this kind of fraction, and it gives a shortcut.

Check it on the example above: $p = 1$, $q = 6$, $r = 2$, $s = 1$ gives $ps - qr = 1 - 12 = -11$. ✓. When a problem has a parameter such as $p$, this turns the question into a simple inequality about $p$. On a test, still show the algebra that leads to it.

Decreasing from some term on

Many sequences go up first and then come down for good. You need to find where the behavior changes. Solve the inequality $a_{n+1} - a_n < 0$ for $n$, then keep the natural numbers that satisfy it.

The same idea handles a parameter. The condition must hold for every $n \ge 1$, so you must find the worst case.

The ratio test for positive terms

For sequences with powers, a ratio is often simpler than a difference.

Some problems call this the quotient criterion. It is the same ratio test.

The same idea works for a ball that rebounds to $70\%$ of its previous height: $h_n = 6(0.7)^n$ feet after a drop from $6$ ft, with ratio $0.7 < 1$. In general, a positive geometric sequence decreases exactly when the common ratio is between $0$ and $1$.

Ratios with powers and factorials

When $a_n$ has a factorial, use $(n+1)! = (n+1)\cdot n!$ so that most of the expression cancels in the ratio.

Powers of $n$ work the same way, but now the ratio has $(n+1)^2$ or similar, so you must solve an inequality.

Radicals, shifted exponentials, and alternating signs

Sometimes the trick is to rewrite $a_n$ in a form where monotonicity is easy to see.

Radicals: multiply by the conjugate

A constant plus a decaying exponential

For $T_n = A + B\,r^n$ with $0 < r < 1$, the constant $A$ cancels in the difference:

$T_{n+1} - T_n = B r^{n+1} - B r^n = B r^n (r - 1)$.

Alternating signs and absolute value

A sequence like $d_n = \dfrac{(-1)^n}{n^2}$ has terms $-1, \tfrac14, -\tfrac19, \tfrac1{16}, \ldots$. It goes down, up, down, up, so it is not monotonic. But its absolute values $e_n = |d_n| = \dfrac{1}{n^2}$ are positive, so the ratio test works: $\dfrac{e_{n+1}}{e_n} = \dfrac{n^2}{(n+1)^2} < 1$. So $(e_n)$ is strictly decreasing.

Bounded below, and how big are the steps?

A strictly decreasing sequence is always bounded above by its first term $a_1$. Whether it is bounded below depends on the sequence. To show a lower bound, rewrite $a_n$ so that a positive piece is added to a constant.

For larger exponents, solve $2^{n+1} > 3000$ with logarithms: $(n+1)\ln 2 > \ln 3000$.

Building new sequences: sums and products

Products of decreasing sequences

If $(a_n)$ and $(b_n)$ are strictly decreasing, then $(a_n + b_n)$ is too: add the inequalities $a_{n+1} < a_n$ and $b_{n+1} < b_n$.

Products are trickier. Multiplying inequalities needs positive numbers.

Partial sums

The partial sum $S_n = a_1 + a_2 + \cdots + a_n$ satisfies $S_{n+1} - S_n = a_{n+1}$. So if every term $a_n$ is negative, then $(S_n)$ is strictly decreasing: you keep adding negative numbers.

Another useful tool is the sum of a geometric series: $\displaystyle\sum_{i=1}^{n} a_1 r^{\,i-1} = a_1\,\frac{1 - r^n}{1 - r}$. For example, $S_n = \sum_{i=1}^n \dfrac{1}{2^i} = 1 - \dfrac{1}{2^n}$. The gap to the limit $1$ is $E_n = 1 - S_n = \dfrac{1}{2^n}$, which is strictly decreasing because its ratio is $\tfrac12 < 1$.

Recurrences: algebra and induction

Some sequences are given by a rule that builds each term from the previous one. There are two main ways to prove they decrease.

Rewrite the difference

For $x_1 = 3$ and $x_{n+1} = \dfrac12\left(x_n + \dfrac{5}{x_n}\right)$, the terms approximate $\sqrt5$ from above. If you know $x_n > \sqrt5$ for all $n$, you can prove that the sequence decreases.

Mathematical induction

When the rule is $a_{n+1} = f(a_n)$ and the sign of the difference is not obvious, prove $a_{n+1} < a_n$ for all $n$ by induction:

1. Base case: check $a₂ < a_1$.

2. Inductive step: assume $a_{n+1} < a_n$ for some $n$ and show that $a_{n+2} < a_{n+1}$, using the recurrence.

3. Conclusion: the statement is true for all $n \ge 1$.

Back to the start: Chloe's cocoa, the car, and the basketball all follow the same rule. A basketball that keeps $80\%$ of its height on every bounce has $h_n = h_0(0.8)^n$. Its terms are positive and $\dfrac{h_{n+1}}{h_n} = 0.8 < 1$, so it is strictly decreasing. The cocoa fits the shifted-exponential pattern, where the difference is $B r^n(r-1) < 0$. And the car's value, which is multiplied by a number between $0$ and $1$ every year, is the ratio test again.

Summary