How to show following estimate with constant depends only on n? We have
$$\frac{k^{1-\frac{2}{n}}}{2k+n-2}\le \frac{k^{1-2/n}}{2k}=\frac{1}{2k^{2/n}}$$
But I am unable to $$\frac{1}{2k^{2/n}}\le c(n)  \frac{1}{(k+1)^{2 / n}}$$
I could not able to show original estimate.
Any help or hint will be greatly appreciated.
 A: We have
$$
\frac{1}{{2k^{2/n} }} = \frac{1}{2}\left( {1 + \frac{1}{k}} \right)^{2/n} \frac{1}{{(k + 1)^{2/n} }} \le \frac{1}{2}e^{2/(kn)} \frac{1}{{(k + 1)^{2/n} }} \le \frac{1}{2}e^{2/n} \frac{1}{{(k + 1)^{2/n} }}.
$$
Note that $\frac{1}{2}e^{2/n}  \le \frac{1}{2}e$ for $n\geq 2$.
A: Here is an alternative approach, using what you have shown so far. To show the desired inequality, it suffices to show that
$$
2k^{2/n} \geq (k+1)^{2/n},
$$
where we will see that we get $c(n)=1$. We have
$$
2k^{2/n} \geq (k+1)^{2/n} \iff (2^n-1)k^2-2k-1 \geq 0.
$$
We define the polynomial $f_n : \mathbb{R} \to \mathbb{R}$ by $f_n(x) := (2^n-1)x^2-2x-1$. For the case $n=2$ we get $f_2(x) = 3x^2-2x-1$. Moreover, we have $f_2(1)=0$, and you can check that $f_2'(1)\geq0$. Hence, we have $f_2(x) \geq 0$ for all $x \geq 1$ and in particular $f_2(k) = 3k^2-2k-1 \geq 0$ for all $k \in \mathbb{N}_{\geq1}$.
If $n > 2$, you can check that we get a zero $x_0$ of $f$ in the interval $(0,1)$ and that $f_n'(x_0) \geq 0$. Therefore, we get $f_n(x)\geq 0$ for all $x \geq x_0$, and in particular $f_n(k) = (2^n-1)k^2-2k-1 \geq 0$, since $k \geq 1 > x_0$. If it helps, you can plot the polynomial $f_n(x)$ for different values of $n \geq 2$, so you get an idea of what is going on here.
Now we have shown that $2k^{2/n} \geq (k+1)^{2/n}$, so we get (by the work you have already done)
$$
\frac{k^{1-2/n}}{2k+n-2} \leq \frac{1}{2k^{2/n}} \leq \frac{1}{(k+1)^{2/n}}.
$$
Now you see that we can indeed simply choose $c(n)=1$.
