Show that $\lim_{n\to\infty} n^{2\alpha-1}-n^{2\alpha-2}\neq0$, if $\alpha>\frac{1}{2}$

Question

I need to show that $$ \lim_{n\to\infty} n^{2\alpha-1}-n^{2\alpha-2}\neq0, $$ if $\alpha>\frac{1}{2}$.

Maybe this is easier to read: $$ \frac{1}{n}n^{2\alpha}-\frac{1}{n^2}n^{2\alpha}. $$ I can sort of argue that $\lim_{n\to\infty}\frac{1}{n}n^{2\alpha}=\infty$, because $2\alpha>0$, so $n^{2\alpha}\geq n$ for each $n\in\mathbb N$. But I can't really squeeze a proof out of this.

So what I basically need is that $n^{\lambda}$, with $\lambda>0$, goes "faster to infinity", than $\frac{1}{n}$ and $\frac{1}{n^2}$ go to zero.

Is there a theorem I could use? Or any other way to look at it?

Answer 1

Let $\alpha>0$. You have $$ n^{2\alpha-1}-n^{2\alpha-2} = n^{2\alpha-1}\left(1-\frac{1}{n}\right) $$ The first factor goes to $\infty$ with $n$, as $2\alpha-1>0$. The second (in parenthesis) converges to $1$. This should be enough for you to conclude directly.

If not, just observe that, for $n\geq 2$, $$ n^{2\alpha-1}\left(1-\frac{1}{n}\right) \geq n^{2\alpha-1}\left(1-\frac{1}{2}\right) = \frac{n^{2\alpha-1}}{2} \xrightarrow[n\to\infty]{}\infty $$

Answer 2

Your first equation does not make sense, because you are subtracting two identical quantities and claiming that their difference is nonzero.

You then write $\frac{1}{n^2}n^{2\alpha}-\frac{1}{n}n^{2\alpha} $. This is just $n^{2\alpha -2} - n^{2\alpha -1}$ which is the same as $n^{2\alpha -1}(\frac1{n} -1) $. The term $(\frac1{n} -1) \to -1$ as $n \to \infty$, so the limit depends on $n^{2\alpha -1}$.

If $2\alpha -1 > 0$, or $\alpha > \frac12$, $n^{2\alpha -1} \to \infty$.

If $2\alpha -1 < 0$, or $\alpha < \frac12$, $n^{2\alpha -1} \to 0$.

If $2\alpha -1 = 0$, or $\alpha = \frac12$, $n^{2\alpha -1} = 1$.