<s>[INST] Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$ I want to check if the following sequence converges:

$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$

I proved the sequence is increasing :
$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$
Now I should prove it is bounded above. let's write some terms of the equation:
\begin{align}
x_0&=1 \\[2ex]
x_1&=1\cdot\left(1+\dfrac{1}{2^1}\right)\\[2ex]
x_2&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\\[2ex]
x_3&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\cdot\left(1+ \dfrac{1}{2^3}\right)\\[2ex]
\end{align}
So, we can write:
$$x_{n+1}=\left(1+\frac{1}{2^1}\right)\cdot\left(1+ \frac{1}{2^2}\right)\cdots\left(1+\frac{1}{2^{n}}\right)\cdot\left(1+\frac{1}{2^{n+1}}\right)$$
Here, I'm not sure how to prove it is bounded above.
 [/INST] You have
$$
x_n = \prod_{k=1}^n \left(1+\frac{1}{2^k}\right)
= e^{\sum_{k=1}^n\ln \left(1+\frac{1}{2^k}\right)}
\leq e^{\sum_{k=1}^n\frac{1}{2^k}}
\leq e^{\sum_{k=1}^\infty\frac{1}{2^k}}
= e
$$
where for the first inequality we used that $\ln(1+x)\leq x$ for all $x>-1$. This shows the sequence is bounded.
</s>