Let $l^2=\left\{(x_n)_n|\, \sum\limits_{n=1}^{\infty}|x_n|^2<\infty\right\}$ with norm $$\|x\|_2'=\left(\sum\limits_{n=1}^{\infty}\frac{n^2}{n^2+1}|x_n|^2\right)^{1/2}$$
So, is it Banach space? I know how to done it with norm $\|.\|_2$ but now I'm confused does $\frac{n^2}{n^2+1}$ change the result in any way?
The linear operator $T:\ell^2\to\ell^2$ defined by $$T(x_n)=\left(\sqrt\frac{n^2}{n^2+1}x_n\right)$$ is bounded (with norm $1$), one-to-one and onto, and thus an isomorphism of Banach spaces. Hence the new norm $\|x\|_2'=\|Tx\|$ is equivalent to the original norm on $\ell^2$, with bounds $$\|T^{-1}\|^{-1}\|x\|\leq\|x\|_2'\leq\|x\|.$$
