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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

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    $\begingroup$ There's not a single word in your "answer"! $\endgroup$ – Lord Shark the Unknown Oct 12 '17 at 6:44
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    $\begingroup$ You have to add a backslash like this \{ and \} in order for the curly braces to appear in math mode, see e.g. here. $\endgroup$ – Flatfoot Oct 12 '17 at 7:07
  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig Oct 12 '17 at 10:48
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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 $$\frac{2n^2}{n^2}.$$

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

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    $\begingroup$ 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 $0$. $\endgroup$ – 5xum Oct 12 '17 at 7:08

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