- #1
JOATMON
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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.
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.
=============
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.
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