Let $(\gamma_n)_{n \geq 1}$ be a sequence of positive real numbers satisfying $$\sum_{n \in \mathbb{N}}\gamma_n = + \infty \text{ and }\sum_{n \in \mathbb{N}}\gamma_n^2 < + \infty$$

I would like to know if the sequence $(u_n)_{n \geq 1}$ defined by $$\forall n \in \mathbb{N},~u_n = \sum_{i=1}^n \gamma_i \prod_{j=i+1}^n (1-\gamma_j)^2$$ is bounded


I have proven that with the same conditions on $(\gamma_n)_{n \geq 1}$, we have $$ \sum_{i=1}^{t} \gamma_i^2 \prod_{j = i+1}^{t} (1-\gamma_j)^2 \underset{n \to \infty}{\longrightarrow} 0$$ using the fact that $\prod_{j = i+1}^{t} (1-\gamma_j)^2 = e^{2 \sum_{j = i+1}^{t} \ln(1-\gamma_j)} \leq e^{- 2 \sum_{j = i+1}^{t} \gamma_j}$


Yes, the sequence is bounded. As you already suggested yourself, it's useful to estimate the product by $\exp \left( -2\sum_{k=j+1}^n \gamma_k \right)$.

The statement is clearest in its continuous version, for integrals: then it's immediately obvious that $$ \int_1^N f(x) e^{-2\int_x^N f(t)\, dt}\, dx = \frac{1}{2}\left( 1 - e^{-2\int_1^N f(t)\, dt}\right) \to \frac{1}{2} . $$

In the original setting, we can mimic the integration by writing $$ e^{-2\sum_{k=j+1}^n \gamma_k} - e^{-2\sum_{k=j}^n \gamma_k} = e^{-2\sum_{k=j+1}^n \gamma_k} \left( 1 - e^{-2\gamma_j} \right) , \quad\quad\quad\quad (1) $$ and $1-e^{-2\gamma_j} = 2\gamma_j + O(\gamma_j^2)$. The error term is summable and can be ignored, so we can approximately replace the summands $\gamma_je^{-2\sum_{k=j+1}^n\gamma_k}$ by ($1/2$ times) the LHS of (1), and then the sum becomes telescoping.

In fact, by keeping track more carefully of what the first step (replacing the product with the exponential) does, we can show that $u_n$ approaches a limit.

  • $\begingroup$ Thank you. I also pasted your answer here. $\endgroup$ – mfrt Aug 17 '18 at 10:16
  • $\begingroup$ @mfrt: Sure, that's fine. $\endgroup$ – Christian Remling Aug 18 '18 at 0:49

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