Is this sequence arithmetic, geometric, or neither?

[Lebombo]

Homework Statement

Is the sequence $\frac{1}{1}, \frac{1}{2}, \frac{1}{3} , \frac{1}{4}...\frac{1}{n}$ arithmetic or geometric?

Homework Equations

Common difference and Common ratio formulas

The Attempt at a Solution

I found the common difference from $a_{2} - a_{1} =d_{1}$ and common difference from $a_{3} - a_{2} =d_{2}$. Since $d_{1}≠ d_{2}$ , then this sequence is not arithmetic.

I did the same thing for the common ratio and found $r_{1}≠ r_{2}$. So this sequence is not geometric either.

It is simply a sequence defined by the sigma notation $\sum_{k=1}^{n} \frac{1}{n}$P.S. I found these topics in an Algebra book, but the topic of sequences and series are also present in my calculus book. To moderators, free to move this to the Algebra section if it's felt that this topic would fit better there.

 

[jedishrfu]

its a harmonic sequence;

http://www.britannica.com/EBchecked/topic/1500010/harmonic-sequence

 

[Lebombo]

Much appreciated.

 

[Mute]

It is simply a sequence defined by the sigma notation $\sum_{k=1}^{n} \frac{1}{n}$

Your expression is incorrect. It should be

$$\sum_{k=1}^n \frac{1}{k}.$$

The variable $k$ is what is being summed over, so it must appear in the sum. Here, n is the upper limit.

As written, $\sum_{k=1}^{n} \frac{1}{n}$ is interpreted as $\frac{1}{n}\sum_{k=1}^n 1 = \frac{1}{n}n = 1$.

You made a similar mistake when writing the sum in your other thread. Be careful, otherwise your equations could easily be misinterpreted.

 

[Lebombo]

Thanks, although I'm still not 100% proficient on the sigma notation, that was a definite typo as opposed to not having the knowledge in a previous thread. Appreciate the correction.

 

[HallsofIvy]

$$\sum_{k=1}\frac{1}{k}$$
is a series. You said this was a sequence which would be denoted as
$$\left\{\frac{1}{k}\right\}_{k=1}^n$$