(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given the sequence

[tex] L_n = Q \cdot (1 + \frac{r}{n}} )^n [/tex]

where [tex] Q > 0, r > 0 [/tex]

show that it either converges or diverges.

2. Relevant equations

Restating the sequence as a recurrence could (maybe) be of help:

[tex] L_{n+1} = L_n \cdot (1 + \frac{r}{n}} ) [/tex] and [tex] L_1 = Q \cdot (1 + \frac{r}{1}}) [/tex]

3. 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 [tex] e^r \cdot Q [/tex], so there exists a [tex] N \in \mathbb{R} [/tex] such that

[tex] \vert Q \cdot (1 + \frac{r}{n}} )^n - e^r \cdot Q \vert < \epsilon [/tex]

for all [tex] n \geq N [/tex].

The problem with this method is, that I can't figure out what to do upon rewriting the above inequality:

[tex] \vert \frac{(n+r)^n}{n^n} - e^r \vert < \frac{\epsilon}{Q} [/tex]

Ideally, I should either solve for [tex]n[/tex] , thus allowing me to find a suitable [tex]\epsilon[/tex] or reduce the inequality to some form where choosing and [tex]\epsilon[/tex] 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:

[tex]r > 0[/tex], so [tex] (1 + \frac{r}{n}}) > 1 [/tex], so [tex] L_{n+1} > L_n [/tex].

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 [tex]Q \cdot (1 + r) < Q \cdot e^r [/tex] 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 [tex]L_k < Q \cdot e^r[/tex], I need to prove that [tex]L_{k+1} < Q \cdot e^r[/tex].

This problem reduces to showing the following:

[tex](1 + \frac{r}{k})^k < e^r [/tex] iff [tex](1 + \frac{r}{k})^{k+1} < e^r[/tex]

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Determining the convergence of a recursive sequence

**Physics Forums | Science Articles, Homework Help, Discussion**