Do These Mathematical Sequences Converge?

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SUMMARY

This discussion focuses on the convergence of three mathematical sequences: 5n/(2n-1), e^n/n, and e^n/(1+e^n). The sequence 5n/(2n-1) converges to 5/2 as n approaches infinity. The second sequence, e^n/n, diverges based on the growth rate test, while the third sequence, e^n/(1+e^n), also diverges when analyzed as a sequence rather than a series. The participants clarify the distinction between sequence convergence and series convergence, emphasizing the importance of limit behavior.

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  • Understanding of limits in calculus
  • Familiarity with sequence and series convergence
  • Knowledge of growth rate tests for sequences
  • Basic integration techniques, specifically antiderivatives
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  • Learn about the Integral Test for series convergence
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Banana Pie
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Which of these converge?

1. 5n/ (2n-1)

2. e^n/ n

3. e^n/ (1+e^n)

Attempt:

1) lim n-> ∞ 5n/(2n-1) = 5n/2n = 5/2 ≠ 0 so diverge?

2) Change n to x
e^x/ (1 + e^x)
Antiderivative: ln |1 + e^x|
lim t->∞ of ln |1 + e^x|
ln |1 + e^∞| - ln |1 + e^0|
Got stuck here

3) Help :(
 
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By converge, do you mean limit exists as n->∞ ? If not, what do you mean? You don't give any information on what limit rules you are familiar with and allowed to use. Please add more such information.
 
Banana Pie said:
Which of these converge?

1. 5n/ (2n-1)

2. e^n/ n

3. e^n/ (1+e^n)

Attempt:

1) lim n-> ∞ 5n/(2n-1) = 5n/2n = 5/2 ≠ 0 so diverge?

2) Change n to x
e^x/ (1 + e^x)
Antiderivative: ln |1 + e^x|
lim t->∞ of ln |1 + e^x|
ln |1 + e^∞| - ln |1 + e^0|
Got stuck here

3) Help :(
Are these sequences of numbers or are they infinite series? The first problem, interpreted as a sequence, converges to 5/2, as you said.
 
I will assume that you are looking for whether or not the sequences converge as n goes to infinity.
As Mark pointed out, you correctly found that the limit of #1 exists as n goes to infinity, so the sequence converges. If you are looking for the sum of the terms, i.e. the series, then your rule about the limit not being equal to zero would be appropriate.
For #2, it looks like you were working on #3.
Also, when you do the integral test, it implies you are looking for the infinite sum, i.e.:
##\lim_{n\to \infty} \sum_{k = 1}^n \frac{e^k}{k}. ##
If that is what you are looking for, then none of these (series) converge.

However, it looks like these are better suited as problems about sequence convergence, so let's go with that.

For #1, the limit exists, so the sequence converges.
For #2, you can try a growth rate test.
##x_n = \frac{e^n}{n}, x_{n+1} = \frac{e^{n+1}}{n+1}.##
##G=\frac{x_{n+1}}{x_n} = \frac{n e^{n+1}}{(n+1)e^n}##
If the limit as n goes to infinity of G is greater than 1, then the sequence is not converging.

For #3, you can apply the same logic you did for #1, and just keep the terms of the highest order.
 

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