Convergence: Epsilon-N Definition

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The discussion focuses on determining the convergence of a sequence using the epsilon-N definition. Participants explore the role of constants like e^2 and e^3, noting their diminishing significance as n approaches infinity. One user successfully applies L'Hospital's Rule, concluding that the sequence converges to 2. They then question whether separating logarithms affects this conclusion and how to properly utilize the epsilon-N definition with L = 2. The conversation emphasizes the importance of understanding logarithmic properties to facilitate the convergence analysis.
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Homework Statement


45068354_253709458645469_3577857059221143552_n.png?_nc_cat=106&_nc_ht=scontent.fhlz2-1.png


Homework Equations


epsilon - N definition of convergence [/B]

The Attempt at a Solution


Not sure how to determine whether this sequence converges or not. Thought it could have something to do with the fact that e^2 and e^3 are just constants, so when n > infinity those e terms become negligible in comparison. If you had to apply convergence definition how could you do that without L? [/B]
 

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UOAMCBURGER said:

Homework Statement


View attachment 232999

Homework Equations


epsilon - N definition of convergence

The Attempt at a Solution


Not sure how to determine whether this sequence converges or not. Thought it could have something to do with the fact that e^2 and e^3 are just constants, so when n > infinity those e terms become negligible in comparison. If you had to apply convergence definition how could you do that without L?
Do you know rules for logarithms ?

Such as ##\ \ln(a\cdot b) =\ln(a)+\ln(b)\,?##

What is ## \ \ln(e^2) \, ?##
 
SammyS said:
Do you know rules for logarithms ?

Such as ##\ \ln(a\cdot b) =\ln(a)+\ln(b)\,?##

What is ## \ \ln(e^2) \, ?##
oh yes i do. Update: I used L'Hospital's Rule to find that the sequence converges to 2. But now separating the logarithms you get 2+ln(n^2)/3+ln(n), so would that change my answer I get using L'Hospital's rule? or can i assume an converges to 2 and then use the definition with L = 2?
 
UOAMCBURGER said:
oh yes i do. Update: I used L'Hospital's Rule to find that the sequence converges to 2. But now separating the logarithms you get 2+ln(n^2)/3+ln(n), so would that change my answer I get using L'Hospital's rule? or can i assume an converges to 2 and then use the definition with L = 2?
Did you mean ##\ \displaystyle \frac{2+\ln(n^2)}{3+\ln(n) } \,?##

If you write a "fraction" all on one line, you need to enclose the numerator and denominator each in parentheses; as in (2+ln(n^2))/(3+ln(n)) .

Also, you may further find that it helps to review rules for logarithms in general.

Another helpful one here is ##\ \displaystyle \ln(a^M) = M\ln(a)\ .##

In addition to this:
You know that the limit is 2, so take the difference between 2 and ##\ \displaystyle \frac{2+\ln(n^2)}{3+\ln(n) } \,.\ ## That should be useful for an ## \epsilon - N\ ## argument. It will also give you what you need to write ##\ \displaystyle \frac{2+\ln(n^2)}{3+\ln(n) } \ ## in a simplified form.
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...

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