Find all values such that the inequality is true

[JOATMON]

Mod note: Moved from a technical math section, so missing the homework template.
This is for an Intro to Analysis course. It's been a very long time since I've taken a math course, so I do not remember much of anything.

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Here is the problem:For the inequality below, find all values n ∈ N such that the inequality is true:

(n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025.============================
Here is my attempt at the problem:Looking at the following set

{n ∈ N: (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}We want to find the lower bound of this set.Suppose A denotes the above set, then we haveA= {n ∈ N: (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}Since the above rational function can be reduced to 1/(2n+1) we have1/(2n+1) <0.025Where we get n>19.5. Since the lower bounds of the set are 19.5, 19.4, 19.3... And so on, this set is bound below by the natural number n=19. Therefore, the above inequality holds true for all n ∈ N greater than or equal to 20.==========
If you can only just tell me what topics I need to review to answer this correctly, I would appreciate it.

 

[ChrisVer]

what is n2,n3?
for example if n=1, n2=1 and n3=170000 you can make the inequality true...
\frac{1+2+3}{2 \times 170000 + 5 + 8 +3} = \frac{6}{340016}= 0.00001764622 &lt; 0.025
and so n=1 is keeping the inequality

 

[JOATMON]

what is n2,n3?

Sorry. I didn't check to see if the format changed once I copied and pasted my problem:
==============================================================

Here is the problem:For the inequality below, find all values n ∈ N such that the inequality is true:

(n² + 2n +3) / (2n³ + 5n²+ 8n + 3) < 0.025.============================
Here is my attempt at the problem:Looking at the following set

{n ∈ N: (n²+ 2n +3) / (2n³ + 5n² + 8n + 3) < 0.025}We want to find the lower bound of this set.Suppose A denotes the above set, then we haveA= {n ∈ N: (n² + 2n +3) / (2n³ + 5n² + 8n + 3) < 0.025}Since the above rational function can be reduced to 1/(2n+1) we have1/(2n+1) <0.025Where we get n>19.5. Since the lower bounds of the set are 19.5, 19.4, 19.3... And so on, this set is bound below by the natural number n=19. Therefore, the above inequality holds true for all n ∈ N greater than or equal to 20. Thus A= {20, 21, 22, 23,...}

 

[mfb]

I don't see the purpose of all those "steps" before reducing the rational function (which you should probably show in more detail).

Where we get n>19.5

Okay (showing the steps wouldn't hurt).

Since the lower bounds of the set are 19.5, 19.4, 19.3... And so on

What is the relevance of those numbers?

Once you have n>19.5 you can directly conclude that n>=20, and write down the set of solutions.

 

[RUber]

One method that might help with the simplification is to flip the inequality.
$\frac{ n^2 +2n+3}{2n^3 + 5n^2 + 8n+3} < .025 \equiv \frac{2n^3 + 5n^2 + 8n+3}{ n^2 +2n+3}>40$
Which as you pointed out can be written as
$2n+1 > 40.$
And you already have the solution.

 

[mfb]

Be careful: to do that you have to check that the two sides cannot be negative. This is easy to do here, but it is a necessary step.