The Past, The Present and The Future!!

We are going to use Stolz-Cesaro theorem here.

$\displaystyle \lim_{n \to \infty} \frac{2016(1^{2015}+2^{2015}+\cdots+n^{2015})-n^{2016}}{2016(1^{2014}+2^{2014}+\cdots+n^{2014}}$

Let $a_n = 2016(1^{2015}+2^{2015}+\cdots+n^{2015})-n^{2016}$ $b_n = 2016(1^{2014}+2^{2014}+\cdots+n^{2014})$

Using Stolz theorem. $\displaystyle \lim_{n \to \infty} \frac{a_n}{b_n}=\lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$

$\displaystyle \lim_{n \to \infty} \frac{2016(n+1)^{2015}-(n+1)^{2016}+n^{2016}}{2016(n+1)^{2014}}$

$\displaystyle \lim_{n \to \infty} n+1-\frac{1}{2016}\left((n+1)^2-n^2(\frac{n}{n+1})^{2014}\right)$

Expand the term using binomial expansion $\displaystyle \lim_{n \to \infty} n+1-\frac{1}{2016}\left(n^2+2n+1-n^2(1-\frac{2014}{n}+\frac{2029015}{n^2}-\cdots)\right)$

$\displaystyle \lim_{n \to \infty} n+1-n+\frac{2029104}{2016}$

$\boxed{\frac{2015}{2}}$

I hope my solution will be comprehensible enough, especially, for those who equip just elementary maths.

First, it's very easy to prove, by induction, that if we have

$\displaystyle \sum _{ j=1 }^{ n }{ { j }^{ k } }$ then the degree of the polynomial obtained(consider as polynomial in n) is k+1.(Note that we're interested in a polynomial of closed form not the form of sum.)

Next, consider the following

$\displaystyle \sum _{ j=1 }^{ n }{ { [(j+1) }^{ k }-{ j }^{ k }] } ={ (n+1) }^{ k }-1=\sum _{ j=1 }^{ n }{ \left( k\cdot { j }^{ k-1 }+\frac { k(k-1) }{ 2 } \cdot { j }^{ k-2 }+... \right) }$

Using this "telescoping" to help prove the very first result and then it's now clear that the leading order and the second leading order is k and k-1, respectively.

Considering only these first two terms, we get

$\displaystyle \frac { 2016({ 1 }^{ 2015 }+...+{ n }^{ 2015 })-{ n }^{ 2016 } }{ 2016({ 1 }^{ 2014 }+...+{ n }^{ 2014 }) } =\frac { 2016(\frac { { n }^{ 2016 }+2016{ n }^{ 2015 } }{ 2016 } -\frac { 2016\cdot 2015 }{ 2\cdot 2016 } \left( \frac { { n }^{ 2015 } }{ 2015 } \right) )-{ n }^{ 2016 } }{ 2016(\frac { { n }^{ 2015 } }{ 2015 } ) } \rightarrow \frac { 2015 }{ 2 }$

note: i have done following assumptions based purely on my observation of pattern 1. coefficient of highest power of the series 1^a +2^a +3^a .. n^a is (1/a+1). 2. coefficient of second highest power of the series 1^a + 2^a + ... n^a is (1/2)

As limit is to infinity,we need to find the ratio of coefficient of highest power of numerator to the coefficient of highest power of denominator.

1^2015 +2^2015 +...n^2015 = (1/2016) n^2016 + (1/2) n^2015 ...

therefore numerator=2016{ (1/2016) n^2016 +(1/2) n^2015 +... } - n^2016 = (1008)*n^2015 +... (n^2016 term gets cancelled)

denominator = 2016{ (1/2015) n^2015 + (1/2) n^2014 .... } = (2016/2015) n^2015 + (1008) n^2014 ... }

therefore answer = (1008)*(2015)/(2016) =2015/2.

The limit we want to calculate is:

$\lim\limits_{n\to0}\frac{2016\sum\limits_1^n k^{2015}-n^{2016}}{2016\sum\limits_1^n k^{2014}}$

We can user Faulhaber's formula to rewrite the sums as polynomials with Bernoulli numbers. The formula states that:

$\sum\limits_1^n k^m = \frac{1}{m+1} \sum\limits_0^m \binom{m+1}{k} B_k^+ n^{m+1-k}$ ,

where $B_k^+$ are the Bernoulli numbers. There exist several approaches to calculate these, but we don't need their explicit formulation, except $B_0^+=1$ and $B_1^+=\frac{1}{2}$ . Applying Faulhaber's formula to our limit, we get:

$\lim\limits_{n\to0}\frac{\sum\limits_0^{2015}\binom{2016}{k}B_k^+ n^{2016-k} - n^{2016}}{\frac{2016}{2015}\sum\limits_0^{2014}\binom{2015}{k}B_k^+ n^{2015-k}} = \frac{2015}{2016}\lim\limits_{n\to0}\frac{\sum\limits_1^{2015}\binom{2016}{k}B_k^+ n^{2016-k}}{\sum\limits_0^{2014}\binom{2015}{k}B_k^+ n^{2015-k}}$ .

The only terms that will survive the limit are $k=1$ in the numerator and $k=0$ in the denominator. In other words:

$\frac{2015}{2016}\lim\limits_{n\to0}\frac{\binom{2016}{1}B_1^+ n^{2015}}{\binom{2015}{0}B_0^+ n^{2015}} = \frac{2015}{2016}\frac{2016}{2}\lim\limits_{n\to0}\frac{n^{2015}}{n^{2015}} = \frac{2015}{2}$ .

i will use Stolz–Cesàro theorem 1

let $b_{n}=2016\left(1^{2014}+2^{2014}+\cdots +n^{2014}\right)$ then $\displaystyle b_{n+1}-b_{n} = 2016(n+1)^{2014}>0, \forall n\in \mathbb{N}$ or $b_{n}<b_{n+1} , \forall n\in \mathbb{N}$ and $\displaystyle \lim_{n\rightarrow \infty} b_{n}=\lim_{n\rightarrow \infty} 2016\left(1^{2014}+2^{2014}+\cdots +n^{2014}\right)=\infty$

So , if $\displaystyle a_{n}=\left(1^{2015}+2^{2015}+\cdots +n^{2015}\right) -n^{2016}$ when $\begin{aligned} \displaystyle \lim_{n\rightarrow \infty} \dfrac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} &=\displaystyle \lim_{n\rightarrow \infty} \dfrac{2016(n+1)^{2015} - \left[ (n+1)^{2016}-n^{2016}\right]}{2016(n+1)^{2014}} \\ &=\displaystyle \lim_{n\rightarrow \infty} \dfrac{\displaystyle 2016\left( n^{2015}+\binom{2015}{1}n^{2014}+\binom{2015}{2}n^{2013}+\cdots+2015n+1\right)-\left(2016n^{2015}+\binom{2016}{2}n^{2014}+\binom{2016}{2}n^{2014}+\cdots+2016n+1 \right ) }{\displaystyle 2016\left( n^{2014}+\binom{2014}{1}n^{2013}+\binom{2014}{2}n^{2012}+\cdots+1\right)} \\ &=\displaystyle \lim_{n\rightarrow \infty} \dfrac{\displaystyle \dfrac{2016\times 2015}{2}n^{2014}+\dfrac{2016\times 2015\times 2013}{2}n^{2013}+\cdots -n }{\displaystyle 2016\left( n^{2014}+\binom{2014}{1}n^{2013}+\binom{2014}{2}n^{2012}+\cdots+1\right)} \\ &=\displaystyle \lim_{n\rightarrow \infty} \dfrac{\frac{2016\times 2015}{2}}{2016}=\frac{2015}{2} \end{aligned}$

we can use Stolz–Cesàro theorem 1 we get

$\displaystyle \lim_{n \to \infty}\, \dfrac{a_{n}}{b_{n}}=\lim_{n \to \infty}\, \dfrac{\displaystyle \left(1^{2015}+2^{2015}+\cdots +n^{2015}\right) -n^{2016}}{2016\left(1^{2014}+2^{2014}+\cdots +n^{2014}\right)}=\lim_{n \to \infty}\, \dfrac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\boxed{\displaystyle \dfrac{2015}{2}=1007.5}$

Let's define $S_p(n)=\displaystyle \sum_{k=1}^{n} k^p$ .

The sums in the limit can be obtained by applying the following recurring formula:

$S_p(n)=\frac{1}{p+1}\left[(n+1)^{p+1}-1-{{p+1} \choose 2} S_{p-1}(n)-...-{{p+1} \choose p} S_{1}(n)-n \right]$

which I derived some time ago and explained in the following document (in Catalan).

We just need to find the first two terms of the formula for the sum in the numerator and only the first one for the sum in the denominator, as the rest of the terms vanish when $n \to \infty$ .

$S_{2015}=\dfrac{n^{2016}}{2016}+\dfrac{n^{2015}}{2}+\ldots$

$S_{2014}=\dfrac{n^{2015}}{2015}+\ldots$

$\displaystyle \lim_{n \to \infty} { \dfrac{ 2016 \left ( \dfrac{n^{2016}}{2016} + \dfrac{n^{2015}}{2}+\ldots \right ) - n^{2016} }{2016 \left ( \dfrac{n^{2015}}{2015}+\ldots \right ) }}=\lim_{n \to \infty} { \dfrac{2016 \cdot \dfrac{n^{2015}}{2} }{2016 \cdot \dfrac{n^{2015}}{2015} } } = \boxed{\dfrac{2015}{2}}$

I used truncated Faulhaber's formula (see https://en.wikipedia.org/wiki/Faulhaber%27s formula) up to second term, i.e. sum (k=1)^n k^p = n^(p+1)/(p+1) + n^p/2 + o(n^p).

We get (here p=2015), Numerator = (p+1) [n^(p+1)/(p+1) + n^p/2 + o(n^p)] - n^(p+1) = (p+1) n^p/2 + o(n^p) and Denominator = (p+1) [n^p/p + n^(p-1)/2 + o(n^(p-1))] = (p+1) n^p/p + o(n^p). Thus, Ratio = Numerator/Denominator = [(p+1) n^p/2 + o(n^p)]/[(p+1) n^p/p + o(n^p)]. For n tends to infinity, the o(.) term can be removed. We get Ratio = p/2. For p=2015, we get ratio = 1007.5.