Evaluating $\lim\limits_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$ Evaluate  $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$

another method, using Stolz–Cesàro theorem:
let ${ x }_{ n }=\left( p+1 \right) \left( { 1 }^{ p }+{ 2 }^{ p }+...+{ n }^{ p } \right) -{ n }^{ p+1 },{ y }_{ n }=\left( p+1 \right) { n }^{ p }$
$$\lim _{ x\rightarrow \infty  }{ \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } = } \lim _{ x\rightarrow \infty  }{ \frac { \left( p+1 \right) { \left( n+1 \right)  }^{ p }-{ \left( n+1 \right)  }^{ p+1 }+{ n }^{ p+1 } }{ \left( p+1 \right) \left( { \left( n+1 \right)  }^{ p }-{ n }^{ p } \right)  } = } \\ =\lim _{ x\rightarrow \infty  }{ \left( \frac { \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right)  }{ 2 } { n }^{ p-2 }+...+1 \right)  }{ \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right)  }{ 2 } { n }^{ p-2 }+...+1-{ n }^{ p } \right)  }  \right) + } \\ +\frac { -{ n }^{ p+1 }-\left( p+1 \right) { n }^{ p }-\frac { p\left( p+1 \right)  }{ 2 } { n }^{ p-1 }-...-1+{ n }^{ p+1 } }{ \left( p+1 \right) \left( { n }^{ p }+p{ n }^{ p-1 }+\frac { p\left( p-1 \right)  }{ 2 } { n }^{ p-2 }+...+1-{ n }^{ p } \right)  } $$
let's cobmine all coefficients of n,then divide numerator and denominator by $n^{ p-1 }$ and define sum of the  all terms  no more -1 power with $o\left( \frac { 1 }{ n }  \right) $
$$\\ \\ \lim _{ x\rightarrow \infty  }{ \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } = } \lim _{ x\rightarrow \infty  }{ \frac { \frac { p\left( p+1 \right)  }{ 2 } +o\left( \frac { 1 }{ n }  \right)  }{ p\left( p+1 \right) +\left( \frac { 1 }{ n }  \right)  } =\frac { 1 }{ 2 }  } \\ $$
