How do we calculate the following limit?

Question

The limit to be found is:-

$$ \lim_{x \to \infty} \frac{2+4+...+2^x}{2^x +1}$$

My attempt:-

Taking $ 2^x $ common from numerator and Denominator,

$$ \lim_{x \to \infty} \frac{(2/2^x)+(4/2^x)+...+1}{1 +(1/2^x)} $$

$$ \frac{(2/2^{\infty})+(4/2^{\infty})+...+1}{1 +(1/2^{\infty})} = 1 $$

So the limit is 1.

Is my answer/attempt correct?

Answer 1

For a sum of $n$ first adds of the geometric progression $a$ with $q_a\neq1$ we have: $$S_n=\frac{a_1(q^n-1)}{q-1}.$$ Id est, it's $$\frac{2(2^x-1)}{2^x+1}=\frac{2\left(1-\frac{1}{2^x}\right)}{1+\frac{1}{2^x}}\rightarrow2.$$

Answer 2

By Stolz-Cesaro we obtain

$$ \frac{(2+4+...+2^{x+1})-(2+4+...+2^x)}{(2^{x+1}+1)-(2^x +1)}=\frac{2^{x+1}}{2^{x+1}-2^x}=\frac{2}{2-1}=2$$

Your way doesn't work because we have

$$\lim_{x \to \infty} \frac{(2/2^x)+(4/2^x)+...+1}{1 +(1/2^x)}=\lim_{x \to \infty} \frac{\sum_{k=1}^x\frac{1}{2^{x-1}}}{1 +(1/2^x)}$$

and the in the limit the numerator is the geometric series $\sum_{k=1}^\infty\frac{1}{2^{x-1}}$.

Answer 3

Your answer is wrong because you consider that all terms in the numerator but the last are $2^k/2^{\infty}$, hence $0$, which is absolutely false. (By the way, why does the last term remain ?)