Find the Limit $\lim_{n \to \infty}\sum_{k=1}^{\infty}\frac{k^{n}}{1+k^{n+2}}$ Find $\lim_{n \to \infty}\sum_{k=1}^{\infty}\frac{k^{n}}{1+k^{n+2}}$
My ideas: let $ n \in \mathbb N$ be constant, looking at $\frac{k^{n}}{1+k^{n+2}}$, we know
$$\frac{k^{n}}{1+k^{n+2}}\leq\frac{k^{n}}{k^{n+2}}=\frac{1}{k^{2}}$$
but this does not help me because $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\pi^{2}/6$
any ideas?

I would do in this way
$$
\eqalign{
  & \sum\limits_{1\, \le \,k} {{{k^{\,n} } \over {1 + k^{\,n + 2} }}}
  = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}\left( {{{k^{\,n + 2} } \over {1 + k^{\,n + 2} }}} \right)}
  = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}\left( {1 - {1 \over {1 + k^{\,n + 2} }}} \right)}  =   \cr 
  &  = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}}  - {1 \over 2} - \sum\limits_{2\, \le \,k} {{1 \over {k^{\,2} }}\left( {{1 \over {1 + k^{\,n + 2} }}} \right)} 
 \;\mathop  \approx \limits^{n \to \infty } \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}}
  - {1 \over 2} - \sum\limits_{2\, \le \,k} {{1 \over {k^{\,2} }}\left( {{1 \over {k^{\,n + 2} }}} \right)}  =   \cr 
  &  = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}}  - {1 \over 2} - \left( {\sum\limits_{1\, \le \,k} {{1 \over {k^{\,n + 4} }}}  - 1} \right)
 = \zeta (2) + {1 \over 2} - \zeta (n + 4) \to \quad \zeta (2) - {1 \over 2} \cr} 
$$
