Find value of $\lim_{n \to \infty} \sum_{k=1}^{101} \frac{(n+k)^{100}}{n^{99}}-101n$ $$ \lim_{n\to\infty} \left(\frac{(n+1)^{100} + (n+2)^{100} + ... + (n+101)^{100}}{n^{99}} - 101n\right) $$
I have tried multiplying n in numerator and denominator in $$ \left(\frac{(n+1)^{100} + (n+2)^{100} + ... + (n+101)^{100}}{n^{99}}\right) $$ so that both numerator and denominator raised to $100$
Another thing I have thought about is multiplying $n^{99}$ with $101n$ and then subtracting $n^{100}$ from each, that is, $$ \left(\frac{(n+1)^{100} - n^{100} + (n+2)^{100} - n^{100} + ... + (n+101)^{100} - n^{100}}{n^{99}}\right) $$ Will that lead to the solution?
 A: Let
$a_n=\sum_{k=1}^{101}(n+k)^{100}-101n^{100}$, then by Stolz-Cesaro theorem,
\begin{align}
\lim_{n\to\infty} \left(\frac{\sum_{k=1}^{101}(n+k)^{100}}{n^{99}} - 101n\right)&=\lim_{n\to\infty}\frac{a_n}{n^{99}}
=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{(n+1)^{99}-n^{99}}\\
&=\lim_{n\to\infty}\frac{(n+102)^{100}-(n+1)^{100}-101((n+1)^{100}- n^{100})}{(n+1)^{99}-n^{99}}\\
&=
\lim_{n\to\infty}\frac{\frac{102\cdot 101\cdot 100\cdot 99}{2}n^{98}+o(n^{98})}{99n^{98}+o(n^{98})}\\
&=\frac{102\cdot 101\cdot 100}{2}=515100.
\end{align}
A: Note that in every binomial of the numerator, only $n^{100}$ and $n^{99}$ terms will be significant in the limit $n \to  \infty$. So its worth computing only these.
$$(n+k)^{100}=n^{100}+100\cdot k \cdot n^{99} + \cdots$$
Therefore on summing,
$$ \lim_{n\to \infty}\dfrac{101n^{100}+100\cdot\tfrac{1}{2}\cdot101\cdot102\cdot n^{99}+o(n^{98})}{n^{99}}-101n=515100$$
A: $$(n+1)^{100} + (n+2)^{100} + \ldots + (n+101)^{100}\sim 101n^{100}+\sum_{k=1}^{101}100k \;n^{99};\;n\to\infty$$
$$\lim_{n\to\infty} \left(\frac{101n^{100}+\sum_{k=1}^{101}100k \;n^{99}}{n^{99}}-101n\right)=\sum_{k=1}^{101}100k =515100$$
