Math Problem Solver

$$\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},...$$This infinite sequence is a harmonic sequence. The n^th term rule for the sequence is $a_n=\frac{1}{n + 2}$, where $n>0$. The recursive formula is $a_n=\frac{a_{n-1}}{a_{n-1} + 1}$, where $n>1$ and $a_{1}=\frac{1}{3}$.

Harmonic sequence is formed by taking the reciprocal of the terms of the arithmetic sequence. The harmonic sequence formula is the reciprocal of the arithmetic sequence formula.

$$\begin{matrix}\frac{1}{3}&&\frac{1}{4}&&\frac{1}{5}&&\frac{1}{6}\\[6pt]3&&4&&5&&6\\[6pt]&+1&&+1&&+1\\[6pt]\end{matrix}$$

The first row is the harmonic sequence, and the second row is the arithmetic sequence.

Explicit Formula

The formula for a harmonic sequence where $a_1$ is the 1^st term, $d$ is the common difference, and $n$ is the term number is $$a_n=\frac{a_1}{d a_1 (n - 1) + 1}$$

Find $a_1$ and $d$: $$\begin{aligned} a_1&=\frac{1}{3} \\ d&=1 \end{aligned}$$

The n^th term rule is:$$\begin{aligned} a_n&=\frac{a_1}{d a_1 (n - 1) + 1} \\ &=\frac{\frac{1}{3}}{\left(1\right) \left(\frac{1}{3}\right) (n - 1) + 1} \\ &=\frac{1}{n + 2} \end{aligned}$$

Recursive Formula

The formula for a harmonic sequence where $a_{n-1}$ is the (n-1)^th term, $d$ is the common difference, and $n>1$ is $$a_n=\frac{a_{n-1}}{d a_{n-1} + 1}$$

Find $d$: $$\begin{aligned} d=1 \end{aligned}$$

The n^th term rule is:$$\begin{aligned} a_n&=\frac{a_{n-1}}{d a_{n-1} + 1} \\ &=\frac{a_{n-1}}{1 a_{n-1} + 1} \\ &=\frac{a_{n-1}}{a_{n-1} + 1} \end{aligned}$$