Homework Help: Is this sequence arithmetic, geometric, or neither?

1. May 30, 2013

Lebombo

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Common difference and Common ratio formulas

3. 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.

Last edited: May 30, 2013
2. May 30, 2013

Staff: Mentor

3. May 30, 2013

Lebombo

Much appreciated.

4. May 31, 2013

Mute

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.

5. May 31, 2013

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.

6. May 31, 2013

HallsofIvy

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

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