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i need help with this exercise i need prove the serie

$\{x_{n}\}=2^{\frac{2^{n}-1}{2^{n}}}$ Converge.

I know for limits, this

$lim_{n\rightarrow\infty}2^{\frac{2^{n}-1}{2^{n}}}=2$

Now, i need prove that result and i finish my exercise.

I try this:

Let $\epsilon>0$, $\exists N\in\mathbb{N}$ such that $n>N$ then $|x_{n}-2|<\epsilon$

Now i want find that $N$

I try this: $|2^{\frac{2^{n}-1}{2^{n}}}-2|=|2^{1-\frac{1}{2^{n}}}-2|=|2\times2^{-\frac{1}{2^{n}}}-2|=|2(2^{-\frac{1}{2^{n}}}-1)|=2|2^{-\frac{1}{2^{n}}}-1|=2|\frac{1}{2^{\frac{1}{2^{n}}}}-1|=2(1-\frac{1}{2^{\frac{1}{2^{n}}}})$ But i'm stuck in that part, can someone help me?

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    Examine first the convergence of $\log_{2}(x_{m})$. Now, $\log_{2}$ is a continuous function on the positive abscissa semiaxis. – 2017-02-08

2 Answers 2

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I'm going to try this a bit different than usual. It is trivial that $2^{(2^n-1)/2^n} > 0$, because the statement $2^n > 0$ holds for all $ n \in \mathbb{R}$. We now show that the sequence is monotonically increasing for positive $n$, i.e. that $$2^{(2^{n+1}-1)/2^{n+1}}>2^{(2^n-1/2)/2^n}$$ It is now clear that $$2^n-1/2 > 2^n - 1 \implies \frac{2^n-1/2}{2^n} > \frac{2^n-1}{2^n}$$ and since $2^n$ is monotonically increasing for $n>0$, so is your function. Since we have $\lim_{n \to \infty} f(n) = 2$, our function must clearly be bounded by $2$ for positive $n$

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    I think you should drop limit. Monotonic [and bounded] sequences in $\mathbb{R}$ converge. – 2018-10-05
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First, simplify your expression: You can write it as $2^{1-2^{-n}}$.

Now, have an idea in mind: where do you think this sequence goes? Well, as $n$ increases, $2^{-n}$ decreases to zero, so $1-2^{-n}$ goes to $1$, so the expression should go to $2^{1} = 2$. This is your candidate.

However, we will deal with this limit differently. Since each term $2^{1-2^{-n}}$ is greater than $1$, it follows that logarithms are valid in this period. Hence, it is enough to prove that $\log_2 {(2^{1-2^{-n}})} \to \log_2 2$ as $n \to \infty$, since the function $y \to 2^y$ is continuous, and $\log_2$ is just the restricted inverse of this function. (I will not justify this point, it is a rather elaborate proof)

Therefore, all we have to do is to prove that $1-2^{-n} \to 1$. However, this is obvious: given $\epsilon>0$, just find $N$ so large that $2^{-N} < \epsilon$. Hence, the convergence of the given sequence (to $2$) follows from this explanation and the italicized point.