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.
On
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 $c(n)=1$.
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 $n\geq 2$.