To prove the given problem, let's carefully analyze and derive the desired result step-by-step. Define the sequence [tex]\((a_n)\)[/tex] such that:
[tex]$
\lim_{n \to \infty} \left[\prod_{k=1}^n \left(a_k+1\right)\right] = \ell, \quad 0 < \ell \leq \infty.
$[/tex]
We are asked to show that:
[tex]$
\sum_{n=1}^{\infty} \left[\frac{a_n}{\prod_{k=1}^n \left(a_k + 1\right)}\right] = 1 - \frac{1}{\ell}.
$[/tex]
### Step-by-Step Solution
1. Define the product sequence [tex]\( P_n \)[/tex]:
Let [tex]\( P_n = \prod_{k=1}^n (a_k + 1) \)[/tex]. Given [tex]\( \lim_{n \to \infty} P_n = \ell \)[/tex], where [tex]\( 0 < \ell \leq \infty \)[/tex].
2. Express [tex]\( P_{n-1} \)[/tex] and [tex]\( a_n \)[/tex]:
Notice that:
[tex]$
P_n = P_{n-1} (a_n + 1).
$[/tex]
For any [tex]\( n \geq 1 \)[/tex], we can write:
[tex]$
(a_n + 1) = \frac{P_n}{P_{n-1}}
\implies a_n = \frac{P_n}{P_{n-1}} - 1.
$[/tex]
3. Rewrite the summation term [tex]\( \frac{a_n}{P_n} \)[/tex]:
Now, we are interested in the term:
[tex]$
\frac{a_n}{P_n} = \frac{\frac{P_n}{P_{n-1}} - 1}{P_n} = \frac{P_n - P_{n-1}}{P_{n-1} P_n}.
$[/tex]
4. Formulate sum expression:
Therefore, the series we need to evaluate is:
[tex]$
\sum_{n=1}^{\infty} \left[\frac{a_n}{P_n}\right] = \sum_{n=1}^{\infty} \left[\frac{P_n - P_{n-1}}{P_{n-1} P_n}\right].
$[/tex]
5. Separate the series:
We can split the fraction:
[tex]$
\frac{P_n - P_{n-1}}{P_{n-1} P_n} = \frac{1}{P_{n-1}} - \frac{1}{P_n}.
$[/tex]
6. Transform the infinite series into a telescoping series:
This simplifies the sum to:
[tex]$
\sum_{n=1}^{\infty} \left(\frac{1}{P_{n-1}} - \frac{1}{P_n}\right).
$[/tex]
7. Recognize the telescoping nature:
The series is telescoping, meaning:
[tex]$
\sum_{n=1}^{\infty} \left(\frac{1}{P_{n-1}} - \frac{1}{P_n}\right) = \left( \frac{1}{P_0} - \frac{1}{P_1} \right) + \left( \frac{1}{P_1} - \frac{1}{P_2} \right) + \dotsb.
$[/tex]
8. Evaluate the limit:
Since [tex]\( P_0 = 1 \)[/tex], it simplifies to:
[tex]$
1 - \lim_{n \to \infty} \frac{1}{P_n} = 1 - \frac{1}{\ell}.
$[/tex]
This results in:
[tex]$
1 - \frac{1}{\ell}.
$[/tex]
Thus, we have shown that:
[tex]$
\sum_{n=1}^{\infty} \left[\frac{a_n}{\prod_{k=1}^n (a_k + 1)}\right] = 1 - \frac{1}{\ell}.
$[/tex]
