Does this limit converge on e?

Question

Playing around with some math in python today I came across what appears to be an interesting pattern:

Starting at n=1 as n approaches positive infinity, take (n+1)^(n+2)/n^(n+1) and get a list of ratios of exponential expressions. At first glance the ratios between the numbers appeared to be converging to something so...

Next, I took the difference between consecutive ratios, e.g. (n+2)^(n+3)/(n+1)^(n+2)-(n+1)^(n+2)/n^(n+1).

The differences appear to be approaching e (2.718...) as n gets larger.

The first few ratios rounded to the third decimal place are...
2^3/1^2 = 8
3^4/2^3 = 10.125
4^5/3^4 = 12.642
5^6/4^5 = 15.259
6^7/5^6 = 17.916
...

With their differences being...
10.125 - 8 = 2.125
12.642 - 10.125 = 2.517
15.259 - 12.642 = 2.617
17.916 - 15.259 = 2.657
...

After the 13th iteration you get 2.711, and it looks like the series will converge on e as it gets arbitrarily large. That or likely positive infinity and my hunch is off!

Can anyone with better knowledge of limits tell me if I've stumbled across a novel way of calculating e or not?

Here's the python code for those curious (first loop stops at 15 because that's all my cheap phone could handle):

import numpy as np

ratios = []
i = 1
while i < 15:
    a = np.power(i,i+1)
    b = np.power(i+1,i+2)
    print(a)
    ratios.append(b/a)
    i+=1
print(ratios)

x=0
diffs = []
while x < len(ratios) - 1:
    temp = ratios[x+1] - ratios[x]
    diffs.append(temp)
    x+=1

print(diffs)

This reminds of the time I thought I discovered a novel formula for exponents about the golden ratio. I didn't, it was already known and I think this may be the case too but my brief search hasn't turned up anything yet.

Thanks!

Answer 1

Mate ,the first sequence which you have is provided is divergent. Since $ \{u_n\}=\frac {(n+1)^{n+2}}{n^{n+1}} = (1+\frac {1}{n})^n \frac{(n+1)^2}{n} \rightarrow \infty$ as $ n\rightarrow \infty$

Answer 2

We have \begin{align*} &\left( {1 + \frac{1}{{n + 1}}} \right)^{n + 2} (n + 2) - \left( {1 + \frac{1}{n}} \right)^{n + 1} (n + 1) \\ & = \left( {1 + \frac{1}{n}} \right)^{n + 1} \left[ {\left( {\frac{{(n + 2)n}}{{(n + 1)^2 }}} \right)^{n + 1} \frac{{(n + 2)^2 }}{{(n + 1)^2 }} - 1} \right](n + 1) \\ &= \left( {1 + \frac{1}{n}} \right)^{n + 1} \left[ {\left( {1 - \frac{1}{n} + \frac{3}{{2n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right)} \right)\frac{{(n + 2)^2 }}{{(n + 1)^2 }} - 1} \right](n + 1) \\ &= \left( {1 + \frac{1}{n}} \right)^{n + 1} \left( {\frac{1}{n} - \frac{3}{{2n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right)} \right)(n + 1) \\ &= \left( {1 + \frac{1}{n}} \right)^{n + 1} \left( {1 - \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = e + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right). \end{align*}

Answer 3

Consider $$a_n=\frac{ (n+1)^{n+2}}{ n^{n+1}}$$ Take logarithms ad use Taylor series for large $n$. You should get $$\log(a_n)=1+\log (n)+\frac{3}{2 n}-\frac{2}{3 n^2}+\frac{5}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ $$a_n=e^{\log(a_n)}=e n+\frac{3 e}{2}+\frac{11 e}{24 n}-\frac{e}{48 n^2}+O\left(\frac{1}{n^3}\right)$$ Continuing with Taylor series $$a_{n+1}-a_n=e -\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$

Answer 4

Let $a_n=(n+1)^{n+2}/n^{n+1}$ so that $a_n/n\to e$. Now $$a_{n+1}-a_n=n\cdot\frac{a_n}{n}\left(\frac{a_{n+1}}{a_n}-1\right)\tag{1}$$ and $$\frac{a_{n+1}}{a_n}=\frac{n+1}{n}\cdot\frac{a_{n+1}/(n+1)}{a_n/n}\to 1$$ and hence from $(1)$ we see that desired limit equals the limit of $$en\log\frac{a_{n+1}}{a_n}\tag{2}$$ We have $$n\log a_{n+1}-n\log(n+1) =n(n+3)\log(1+(n+1)^{-1}) $$ and $$n\log a_n-n\log n=n(n+2)\log(1+n^{-1})$$ and hence $$n\log\frac{a_{n+1}}{a_n}=n\log(1+n^{-1})+n^2\log(1-(n+1)^{-2})+3n\log(1+(n+1)^{-1})-2n\log(1+n^{-1})$$ and the right hand side tends to $1-1+3-2=1$. The desired limit is thus $e$.