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?
 A: Note that
$${1\over k^2}-{k^n\over1+k^{n+2}}={1\over k^2(1+k^{n+2})}$$
and, for $k\ge2$,
$${1\over k^2(1+k^{n+2})}\le{1\over2^nk^2}$$
so that
$$\sum_{k=1}^\infty{1\over k^2(1+k^{n+2})}={1\over2}+\sum_{k=2}^\infty{1\over k^2(1+k^{n+2})}\le{1\over2}+{1\over2^n}\sum_{k=2}^\infty{1\over k^2}\to{1\over2}+0={1\over2}$$
as $n\to\infty$. It follows that
$$\sum_{k=1}^\infty{k^n\over1+k^{n+2}}=\sum_{k=1}^\infty{1\over k^2}-\sum_{k=1}^\infty{1\over k^2(1+k^{n+2})}\to\sum_{k=1}^\infty{1\over k^2}-{1\over2}={\pi^2\over6}-{1\over2}$$
The key step, really, is to realize that the inequality $2^n\le1+k^{n+2}$ is not satisfied for $k=1$, but is satisfied for $k\ge2$, so the $k=1$ term needs to be split off from the sum.
A: HINT
Look at the summation for a couple of fixed integer $n$. For example, if $n = 5$ you have
$$
\frac{k^5}{1+k^7} \approx \frac{1}{k^2}...
$$
A: I would do in this way
$$
\eqalign{
  & \sum\limits_{1\, \le \,k} {{{k^{\,n} } \over {1 + k^{\,n + 2} }}}
  = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}\left( {{{k^{\,n + 2} } \over {1 + k^{\,n + 2} }}} \right)}
  = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}\left( {1 - {1 \over {1 + k^{\,n + 2} }}} \right)}  =   \cr 
  &  = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}}  - {1 \over 2} - \sum\limits_{2\, \le \,k} {{1 \over {k^{\,2} }}\left( {{1 \over {1 + k^{\,n + 2} }}} \right)} 
 \;\mathop  \approx \limits^{n \to \infty } \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}}
  - {1 \over 2} - \sum\limits_{2\, \le \,k} {{1 \over {k^{\,2} }}\left( {{1 \over {k^{\,n + 2} }}} \right)}  =   \cr 
  &  = \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}}  - {1 \over 2} - \left( {\sum\limits_{1\, \le \,k} {{1 \over {k^{\,n + 4} }}}  - 1} \right)
 = \zeta (2) + {1 \over 2} - \zeta (n + 4) \to \quad \zeta (2) - {1 \over 2} \cr} 
$$
A: The series equals
$$\frac{1}{2}+ \sum_{k=2}^{\infty}\frac{1}{k^{-n} + k^2}.$$
For each $k\ge 2,$ the terms in the last series increase to $1/k^2$ as $n\to \infty.$ By the monotone convergence theorem, the desired limit is
$$\frac{1}{2} +\sum_{k=2}^{\infty}\frac{1}{k^2} = \frac{1}{2}+\left (\frac{\pi^2}{6} - 1\right) = \frac{\pi^2}{6} - \frac{1}{2}.$$
