<s>[INST] With or without squeeze theorem I am doing a proof involving finding the sum of a series $\sum_{x\geq1} \frac{x}{2^x} - \frac{x+1}{2^{x+1}}$ and I need to show that $\lim_{x\to\infty} \frac{x+1}{2^x}$ converges to $0$ without using L'Hopital's theorem.
My attempt:
$\lim_{x\to\infty} \frac{x+1}{2^x} = \lim_{x\to\infty} \frac{x}{2^x} + \lim_{x\to\infty} \frac{1}{2^x}$
We then get that $2^x > x^2 \implies \frac{x}{2^x} < \frac{x}{x^2} = \frac{1}{x}$ and also, $2^x > x \implies \frac{1}{2^x} < \frac{1}{x}$ for sufficiently large $x$. We have that
$$0 \leq \lim_{x\to\infty} \frac{x}{2^x} \leq \lim_{x\to\infty} \frac{1}{x} = 0$$
$$0 \leq \lim_{x\to\infty} \frac{1}{2^x} \leq \lim_{x\to\infty} \frac{1}{x} = 0$$
(Notice the change from $<$ to $\leq$ above. Is that allowed? Also, the series starts at $1$, not $0$.)
Using squeeze theorem, $\lim_{x\to\infty} \frac{x+1}{2^x} = 0$.
Is this correct? Notice my bracketed comment. Any assistance much appreciated.
 [/INST] Since you are in the context of a series, $x$ is natural.
So, you can apply the binomial formula to $2^x = (1+1)^x=\sum_{n=0}^x\binom xn$:
$$0\leq \frac{x}{2^x} = \frac{x}{(1+1)^x} \stackrel{x\geq 2}{\leq} \frac{x}{\binom x2} = \frac 2{x-1}\stackrel{x\to \infty}{\longrightarrow}0$$
Similarly for the case $\frac 1{2^x}$.
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