Relating a product to an exponential

Question

Based on numerical experiments, I think that

$\prod_{i=1}^t \frac{1}{1 + \delta i} = \exp\left(\eta t \right)$

where $\eta = \frac{\log\left(\prod_{i=1}^T \frac{1}{1 + \delta i}\right)}{T}$ for any $T>t$, for integer $T,t$ and real $\delta > 0$.

Here is a numerical example using Python:

import numpy as np

# define product 
def prodf(delta, T):
    prod = 1
    for n in range(T):
        prod *= 1 / (1 + delta)
    return prod

# set inputs
T = 2000
t = 40
delta = 2e-3

# set exponential parameter
eta = np.log(prodf(delta, T)) / T

# compare prediction of product and exponential
print('product at t: ', prodf(delta, t))
print('exponential at t:', np.exp(eta * t))

 # product at t:  0.9231901003222546
 # exponential at t: 0.9231901003222545

Here is my best attempt to prove this:

$\prod_{i=1}^t \frac{1}{1 + \delta i} = \frac{1}{1+\delta} \times \frac{1}{1+2\delta} \times \dots \frac{1}{1+t\delta}$.

Each term in the sequence can be defined as $s_i=\frac{1}{1+i \delta}$.

There exists a number $\mu$ such that $\mu = \frac{1}{T} \sum_{i=1}^T s_i$.

So, the product can be rewritten as $\prod_{i=1}^T \mu$, which is, by definition, $\mu^T$.

Then, we can reasonably say $\mu^T=\exp(\delta T)$, where $\delta = \log(\mu^T)/T$.

Was I successful in my proof? Is the proposed equality true? Was there a more straight-forward way to prove this?

Answer 1

Too long for a comment:

Your code answers my question: when $\eta$ is defined by $$\eta T=\log(\prod_{i=1}^T\frac{1}{1+\delta})$$ for some fixed positive integer $T$ then you found out that $$\eta t=\log(\prod_{i=1}^t\frac{1}{1+\delta})$$ for all integers $0<t<T$. To show this it is enough to prove that $$ \eta =\log\frac{1}{1+\delta}. $$ This is easy to see when you consider that $\eta$ is the arithmetic mean of $T$ equal terms $$ \log\frac{1}{1+\delta} $$ hence it must be equal to this.