Boundedness Of Abundance Of Product Of $N$ Consecutive Integers

Question

Is the abundance of the product of 2 consecutive integers (pronic numbers) bounded? What about the abundance of the product of 3 consecutive integers? What about $n$ consecutive integers? I wrote the following little Python script along the following lines to try and understand it better for the case $n=3$ (and I can generalize it), but I know that this is not a substitute for a proof. I found an abundance of approximately 3.294 for $54\times 55\times 56$, but nothing higher.

import math 
import time

i=1

def abundance(n): 
    sum=0
    i=1

    while i<=(math.sqrt(n)): 
        if n%i==0: 
            if n/i==i: 
                sum=sum+i 
            else:
                sum=sum+i 
                sum=sum+(n/i) 
        i=i+1

    sum=sum-n 
    return sum/n

while True:
  print("Abundance of "+str(i)+", "+str(i+1)+", and "+str(i+2)+" is "+str(abundance(i*(i+1)*(i+2))))
  i+=1
  time.sleep(0.5)

Answer 1

By the abundance of a positive integer $n$, I understand you to mean $\frac{\sigma(n)}{n}$ where $\sigma$ is the sum-of-divisors function.

Since the abundance of single integers is unbounded, so is the abundance of products of two (or more) consecutive integers, because in general: $\frac{\sigma(mn)}{mn}\ge \frac{\sigma(n)}{n}$

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As pointed out by @Daniel Fischer (thank you!), OP is defining abundance where the sum of factors only includes proper factors (i.e., $n$ is excluded in the sum of factors of $n$). I am using the sum of all factors, including $n$. This means that my value of the abundance of an integer $n$ will always be exactly one greater than OP's value. However, the unboundedness argument I give is still valid.