A calculus problem by Akshat Sharda

Assuming $k$ as a positive integer, one can write:

$\large \displaystyle 1^{k+1} = (0+1)^{k+1} = C_{k+1}^0 \cdot 0^{k+1} + C_{k+1}^1 \cdot 0^k \cdot 1 + C_{k+1}^2 \cdot 0^{k-1} \cdot 1^2 + ... + C_{k+1}^{k+1} \cdot 1^{k+1}$

$\large \displaystyle 2^{k+1} = (1+1)^{k+1} = C_{k+1}^0 \cdot 1^{k+1} + C_{k+1}^1 \cdot 1^k \cdot 1 + C_{k+1}^2 \cdot 1^{k-1} \cdot 1^2 + ... + C_{k+1}^{k+1} \cdot 1^{k+1}$

$\large \displaystyle 3^{k+1} = (2+1)^{k+1} = C_{k+1}^0 \cdot 0^{k+1} + C_{k+1}^1 \cdot 2^k \cdot 1 + C_{k+1}^2 \cdot 2^{k-1} \cdot 1^2 + ... + C_{k+1}^{k+1} \cdot 1^{k+1}$

$\large \displaystyle \vdots$

$\large \displaystyle (n+1)^{k+1} = C_{k+1}^0 \cdot n^{k+1} + C_{k+1}^1 \cdot n^k \cdot 1 + C_{k+1}^2 \cdot n^{k-1} \cdot 1^2 + ... + C_{k+1}^{k+1} \cdot 1^{k+1}$

Summing it all up, one ends up with:

$\large \displaystyle (n+1)^{k+1} = (k+1) \sum_{j=1}^{n} j^k + O(n^{k})$

$\color{#20A900} \boxed{ \large \displaystyle \sum_{j=1}^{n} j^k = \frac{(n+1)^{k+1}}{k+1} + O(n^{k}) }$

So:

$\large \displaystyle \lim_{n \rightarrow \infty} \frac{ (n+1)^{1-k} \sum_{j=1}^{n} j^k }{\sum_{j=1}^{n}(nk + j) }$

$\large \displaystyle = \lim_{n \rightarrow \infty} \frac{ (n+1)^{1-k} \cdot \left [ \frac{(n+1)^{k+1}}{k+1} + O(n^{k}) \right ] }{nk\sum_{j=1}^{n}1+ \sum_{j=1}^{n}j }$

Since $n$ goes to $\infty$ :

$\large \displaystyle = \lim_{n \rightarrow \infty} \frac{ (n+1)^{1-k} \cdot \left [ \frac{(n+1)^{k+1}}{k+1} \right ] }{n^2 k + \frac{n^2 + n}{2} }$

$\large \displaystyle = \lim_{n \rightarrow \infty} \frac{2n^2 + 4n + 2}{(2k^2 + 3k+ 1)n^2 + (k+1)n}$

$\color{#20A900} \boxed{ \large \displaystyle = \frac{2}{2k^2 + 3k + 1} }$

Since it equals $\frac{1}{60}$ :

$\large \displaystyle 2k^2 + 3k + 1 = 120$

$\large \displaystyle 2k^2 + 3k - 119 = 0$

$\color{#20A900} \boxed{ \large \displaystyle k = 7}$ or $\color{#20A900} \boxed{ \large \displaystyle k = -8.5 }$

Since we assumed $k$ as a positive integer:

$\color{#3D99F6} \boxed{ \large \displaystyle k = 7}$

Notice that $\displaystyle \lim_{n\to \infty} \dfrac{ (n+1)^{1-k} \displaystyle \sum^{n}_{j=1} j^k}{\displaystyle \sum^{n}_{j=1} (nk+j)}$

$=\displaystyle \lim_{n\to \infty} \dfrac{ n^{k+1}/(n+1)^{k-1} \displaystyle \sum^{n}_{j=1} 1/n( j/n)^k}{n^2 \left[ \displaystyle \sum^{n}_{j=1} 1/n(j/n) \right]+n^2k}$

$=\frac{\int^{1}_{0} x^kdx}{\int^{1}_{0}xdx + k} \times \lim_{n \to \infty} (n/(n+1))^{k+1}=1/60$