Proof that a sequence is convergent I'm asked to prove the convergence of the sequence $$X_n=\left(1+\frac12\right)\left(1+\frac14\right)\left(1+\frac18\right)\cdots\left(1+\frac{1}{2^n}\right)$$ 
I proved that it is increasing through the ratio test and then I want to prove that it is bounded.
My question is the following : is it enough to say that $X_n<\left(\frac32\right)^n$ for $n>1$ or not ? Namely, can we say that a sequence is bounded if $\forall$  $n$ ,  $\exists$ a $N$ such that $X_n<N$ or do I have to find a N such that $\forall$  $n$ , $X_n<N$?
In other words, can a bound depend on $n$ ?
Thanks for your help.
 A: $\ln(X_n) = \displaystyle \sum_{k=1}^n \ln(1+\frac{1}{2^k})$. And $0 < \ln(1 + \dfrac{1}{2^k}) < \dfrac{1}{2^k}$, thus by comparison test the former series converges, and therefore the sequence $\ln(X_n)$ converges which implies $X_n$ converges.
A: It is not enough to say that $X_n<1.5^n$. You need to show that your sequence is bounded, that is, there is a number $C>0$ such that $|X_n|<C$ for any $n$.
For instance, you can note that $1+1/2^k\leqslant (1+1/2^n)^{2^{n-k}}$, and then $X_n<e$.
A: No, the bound cannot depend on $n$.  If it could, then the series 
$$
1, 2, 3, 4, 5, \cdots (x_n = n
$$
would be bounded, choosing $N(n) = n+1$.
One way of showing your product $X$ is bounded is to take the log of $X_n$:
$$
\log X_n = \sum_{k=0}^n \log(1+2^{-k})
$$
Then use the easily proven fact that for $0 < x < 1$
$$
\log(1+x) < x$$
This lets you show that for all $n$, 
$$
\log(X_n) <  \sum_{k=0}^n 2^{-k}) < 2
$$
But a much better way is to prove by induction that 
$$
X_n = 2 -2^{-(2^n-1)}$$
This closed-form expression for $X_n$ is clearly bounded above by $2$.
A: $$X_n=\left(1+\frac12\right)\left(1+\frac14\right)\left(1+\frac18\right)\cdots\left(1+\frac{1}{2^n}\right)$$
$$= \frac{(1-\frac12)(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n})}{1-\frac12}$$
$$=2(1-\frac{1}{2^2})(1+\frac{1}{2^2})\dots (1+\frac{1}{2^n})$$
$$\bbox[5px,border:2px solid red]{X_n=2(1-\frac{1}{2^{2n}})}$$
As $n \to \infty$;
$X_n$ convergers to $2$
$$\bbox[5px,border:2px solid red]{X_n=2}$$
