dane502
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Homework Statement
Given the sequence
L_n = Q \cdot (1 + \frac{r}{n}} )^n
where Q > 0, r > 0
show that it either converges or diverges.
Homework Equations
Restating the sequence as a recurrence could (maybe) be of help:
L_{n+1} = L_n \cdot (1 + \frac{r}{n}} ) and L_1 = Q \cdot (1 + \frac{r}{1}})
The Attempt at a Solution
I've tried two different approaches:
- Using the epsilon-definition of convergence: Using Maple I've found the limit of the sequence to be e^r \cdot Q, so there exists a N \in \mathbb{R} such that
\vert Q \cdot (1 + \frac{r}{n}} )^n - e^r \cdot Q \vert < \epsilon
for all n \geq N.
The problem with this method is, that I can't figure out what to do upon rewriting the above inequality:
\vert \frac{(n+r)^n}{n^n} - e^r \vert < \frac{\epsilon}{Q}
Ideally, I should either solve for n , thus allowing me to find a suitable \epsilon or reduce the inequality to some form where choosing and \epsilon would be obvious. But it seems impossible due to the exponentiation. - Using that a sequence converges iff it is monotonous and has a supremum/infimum
Therefore I've thought of restating the sequence as a recurrence and showing that it is monotunous and has a supremum.
Showing that the sequence is monotonous is easy:
r > 0, so (1 + \frac{r}{n}}) > 1, so L_{n+1} > L_n.
The second part seems unsolvable, though.
I've tried to use induction to show that the sequence has a supremum.
Base step:
I guess that proving Q \cdot (1 + r) < Q \cdot e^r is easy, and so I haven't yet spent time on proving it.
What I have spent time on, however, is the inductive step:
Assuming that L_k < Q \cdot e^r, I need to prove that L_{k+1} < Q \cdot e^r.
This problem reduces to showing the following:
(1 + \frac{r}{k})^k < e^r iff (1 + \frac{r}{k})^{k+1} < e^r
I have no idea how one would go about proving that the left side of the 'iff' implies the right side of the 'iff'
I would greatly appreciate if anyone could help me making some advances using either method 1 or method 2 - or an entirely different method.
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