How to tell if a series is decreasing?

To determine if the terms of a series are decreasing, you need to verify if each term in the sequence is smaller than the term that precedes it. More formally, a sequence of terms, say $a_n$, is considered decreasing (or strictly decreasing) if for every positive integer $n$, the condition $an > a{n+1}$ holds true. This means that as you progress through the terms of the series, their values are consistently getting smaller.

Understanding whether the terms of a series are decreasing is crucial in calculus, especially when applying various convergence tests like the Integral Test or the Alternating Series Test.

Understanding Decreasing Terms

A sequence ${a_n}$ is decreasing if $a_1 \ge a_2 \ge a_3 \ge \dots \ge an \ge a{n+1} \ge \dots$.
It is strictly decreasing if $a_1 > a_2 > a_3 > \dots > an > a{n+1} > \dots$.
The core definition for strict decrease, as highlighted, is $an > a{n+1}$ for all $n$.

Methods to Check if Series Terms Are Decreasing

There are several effective methods to determine if the terms of a sequence $a_n$ (which form a series) are decreasing.

1. Direct Comparison Method

This is the most straightforward method, directly applying the definition.

• Approach: Compare $an$ with $a{n+1}$. If $an > a{n+1}$ for all $n$ (or for all $n$ beyond a certain point), then the sequence is decreasing.
• Example: Consider the terms $a_n = \frac{1}{n}$.
  • We need to check if $an > a{n+1}$, i.e., $\frac{1}{n} > \frac{1}{n+1}$.
  • Since $n+1 > n$ for all $n \ge 1$, it logically follows that $\frac{1}{n} > \frac{1}{n+1}$.
  • Conclusion: The terms of the series $\sum \frac{1}{n}$ are decreasing.

2. Difference Method

This method involves subtracting consecutive terms and checking the sign of the result.

• Approach: Calculate the difference $an - a{n+1}$. If $an - a{n+1} > 0$ for all $n$ (or for $n$ beyond a certain point), the sequence is decreasing.
• Example: Consider the terms $a_n = \frac{n}{n^2+1}$.
  • Calculate $an - a{n+1}$:
    $an - a{n+1} = \frac{n}{n^2+1} - \frac{n+1}{(n+1)^2+1}$
    $= \frac{n}{n^2+1} - \frac{n+1}{n^2+2n+2}$
    To combine, find a common denominator:
    $= \frac{n(n^2+2n+2) - (n+1)(n^2+1)}{(n^2+1)(n^2+2n+2)}$
    Numerator: $n^3+2n^2+2n - (n^3+n^2+n+1) = n^2+n-1$.
  • So, $an - a{n+1} = \frac{n^2+n-1}{(n^2+1)(n^2+2n+2)}$.
  • For $n \ge 1$, the numerator $n^2+n-1$ is positive (e.g., for $n=1$, $1^2+1-1=1>0$; for $n=2$, $2^2+2-1=5>0$). The denominator is also always positive.
  • Conclusion: Since $an - a{n+1} > 0$, the terms of the series $\sum \frac{n}{n^2+1}$ are decreasing.

3. Ratio Method (for Positive Terms)

This method is particularly useful when terms involve factorials or exponentials. It applies only when all terms are positive.

• Approach: Calculate the ratio $\frac{a_{n+1}}{an}$. If $\frac{a{n+1}}{a_n} < 1$ for all $n$ (or for $n$ beyond a certain point), the sequence is decreasing.
• Example: Consider the terms $a_n = \left(\frac{1}{2}\right)^n$.
  • Calculate $\frac{a_{n+1}}{an}$:
    $\frac{a{n+1}}{a_n} = \frac{(1/2)^{n+1}}{(1/2)^n} = \frac{(1/2)^n \cdot (1/2)}{(1/2)^n} = \frac{1}{2}$.
  • Since $\frac{1}{2} < 1$.
  • Conclusion: The terms of the series $\sum \left(\frac{1}{2}\right)^n$ are decreasing.

4. Derivative Method (for Continuous Functions)

If the terms $a_n$ can be represented by a continuous, differentiable function $f(x)$ such that $f(n) = a_n$, you can use calculus.

• Approach: Define a function $f(x)$ such that $f(n) = a_n$. If the first derivative $f'(x) < 0$ for all $x \ge N$ (for some integer $N$), then the sequence $a_n$ is decreasing for $n \ge N$.
• Example: Consider the terms $a_n = \frac{1}{\ln(n)}$ for $n \ge 2$.
  • Let $f(x) = \frac{1}{\ln(x)}$.
  • Find the derivative $f'(x)$:
    Using the chain rule, $f(x) = (\ln(x))^{-1}$, so $f'(x) = -1 \cdot (\ln(x))^{-2} \cdot \frac{1}{x} = -\frac{1}{x(\ln(x))^2}$.
  • For $x \ge 2$, $x$ is positive, and $\ln(x)$ is positive (so $(\ln(x))^2$ is positive).
  • Therefore, $f'(x) = -\frac{1}{\text{positive value}}$ is always negative ($<0$).
  • Conclusion: Since $f'(x) < 0$ for $x \ge 2$, the terms of the series $\sum \frac{1}{\ln(n)}$ are decreasing for $n \ge 2$.

Summary Table of Methods

By applying one or more of these methods, you can confidently determine whether the terms of a given series are decreasing.