Increasing, Decreasing, Monotonic…

Is this right? I'm trying to learn by following some examples in my lecture slides but really want to confirm with someone, preferably someone learned in math, so I thought of posting here. Cheers.

Q: Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded $$\{\frac{2n^2-1}{n}\}^\infty_{n=2}$$

A: $$a_n=\frac{2n^2-1}{n}$$ $$a_{n+1}=\frac{2(n+1)^2-1}{n+1}$$ $$a_{n+1}-a_n$$ $$\frac{2(n+1)^2-1}{n+1}-\frac{2n^2-1}{n}$$ $$\frac{2n^2+4n+2-1}{n+1}-\frac{2n^2-1}{n}$$ $$\frac{2n^2+4n+1}{n+1}-\frac{2n^2-1}{n}$$ $$\frac{2n^3+4n^2+n-2n^3-2n^2+n+1}{n^2+n}$$ $$\frac{2n^3+4n^2+n-2n^3-2n^2+n+1}{n^2+n}$$ $$\frac{2n^2+2n+1}{n^2+n}$$ $$\frac{2n^2}{n^2}$$ $$2$$ Strictly Increasing & Striclty monotonic

• There's not a single word in your "answer"! – Lord Shark the Unknown Oct 12 '17 at 6:44
• You have to add a backslash like this \{ and \} in order for the curly braces to appear in math mode, see e.g. here. – Flatfoot Oct 12 '17 at 7:07
• There should be equal signs connecting every line in your calculation of $a_{n + 1} - a_n$. You are asserting that each line is equal to the previous line and, therefore, equal to $a_{n + 1} - a_n$. However, the assertion $$\frac{2n^2 + 2n + 1}{n^2 + n} = \frac{2n^2}{n^2}$$ is false. – N. F. Taussig Oct 12 '17 at 10:48

The one biggest flaw in your argument is that it contains only equations. Proofs should always tell a story, not just list equations. If you use an equation and you don't explain why you used it, the equation is useless.

That said, there are also calculation flaws, since I don't really see how you can get from $$\frac{2n^2+2n+1}{n^2+n}$$ to $$\frac{2n^2}{n^2}.$$

• oh right sorry. I got carried away while solving and thought I was computing the limit. I've edited it.....is this now correct? – Sean Paul Oct 12 '17 at 7:06

Ok taking your responses into account I have come up with this:

A:

Testing the difference of successive terms,

Let $$a_n=\frac{2n^2-1}{n}$$

So, $$a_{n+1}=\frac{2(n+1)^2-1}{n+1}$$ $$a_{n+1}-a_n$$ $$\frac{2(n+1)^2-1}{n+1}-\frac{2n^2-1}{n}$$ $$\frac{2n^2+4n+2-1}{n+1}-\frac{2n^2-1}{n}$$ $$\frac{2n^2+4n+1}{n+1}-\frac{2n^2-1}{n}$$ $$\frac{2n^3+4n^2+n-2n^3-2n^2+n+1}{n^2+n}$$ $$\frac{2n^3+4n^2+n-2n^3-2n^2+n+1}{n^2+n}$$ $$\frac{2n^2+2n+1}{n^2+n} > 0$$ for all n >= 2

Thus, $\{\frac{2n^2-1}{n}\}^\infty_{n=2}$ is Strictly Increasing & Strictly monotonic

• Rather than posting an answer, you should edit your original question. And your answer could still benefit from a little more words in it. And some explanation why the final expression is greater than $0$. – 5xum Oct 12 '17 at 7:08