Can we expect to find $r,$ large enough, so, $\sum_{n\in \mathbb Z} \frac{(1+n^{2})^{s}}{1+(n-y)^{r}}\leq C (1+y^{2})^{s} $ for all $y\in \mathbb R$? Fix $y\in \mathbb R$ and $s>1.$
Consider the series:
$$I(y)=\sum_{n\in \mathbb Z} \frac{(1+n^{2})^{s}}{1+(n-y)^{r}}.$$

My Question is: Can we expect to find $r$ large enough, so that 
  $$I(y)\leq C (1+y^{2})^{s}$$ 
  where $C$ is some constant; for all $y\in \mathbb R$ ?

Note. See the related question and its answer and also this.
 A: 1) Let $\displaystyle a=\frac{1}{2}\in ]0,1[$ , $s>0$ fixed, and $r$ a large even integer(say $r\geq 2s+2$), $b=a^r$, and  put:
$$I_b(y)=\sum_{n\in \mathbb{Z}}\frac{(1+n^2)^s}{1+b(n-y)^r}$$
Suppose now that $y=m\in \mathbb{Z}$. Then $n\to n-m$ is a bijection of $\mathbb{Z}$ to $\mathbb{Z}$, so we can write:
$$I_b(m)=\sum_{k\in \mathbb{Z}}\frac{(1+(k+m)^2)^s}{1+b(k)^r}$$
We have $$\frac{1+(k+m)^2}{1+m ^2}=1+\frac{k^2}{1+m^2}+2\frac{km}{1+m^2}\leq 1+k^2+|k|\leq 2(1+k^2)$$
hence
$$I_b(m)\leq 2^s(1+m^2)^s \sum_{k\in \mathbb{Z}}\frac{(1+k^2)^s}{1+bk^r}\leq C_1(1+m^2)^s$$
where $C_1$ is a constant depending on $a,s$ and $r$.
2) Now let $y\in \mathbb{R}$, and
$$I(y)=\sum_{n\in \mathbb{Z}}\frac{(1+n^2)^s}{1+(n-y)^r}$$
Put $y=m+x$, with $m\in \mathbb{Z}$ and $x\in [0,1[$.
Suppose that $|n-m|\geq 2$. Then (recall that $a=1/2$):
$$|n-y|=|n-m-x|\geq |n-m|-1\geq a|n-m|$$
Hence with $b=a^r$:
$$I(y)\leq \sum_{|n-m|\leq 1}\frac{(1+n^2)^s}{1+(n-y)^r}+\sum_{|n-m|\geq 2}\frac{(1+n^2)^s}{1+b(n-m)^r}=A+B$$
Now as if $|n-m|\leq 1$, we have $|n|\leq |m|+1$, and $1+(n-y)^r\geq 1$; there exists a constant $C_2$ such that we have $A\leq C_2(1+(|m|+1)^2)^s$. We have also $B\leq I_b(m)$, hence $B\leq  C_1(1+m^2)^s$. Thus there exists a constant $C_3$ such that  $I(y)\leq C_3(1+m^2)^s$. Now there exists a constant $C_4$ such that $1+m^2\leq C_4(1+y^2)$, and we are done. 
