Another limit of sums

A direct application of Stolz–Cesàro theorem should do the trick.

Let $\displaystyle a_n =\sum_{k=1}^n (k+2015)^{2014}$ and $b_n = n^{2015}$ . Because $b_n$ is a strictly monotone and divergent sequence, then Stolz–Cesàro theorem is applicable if $\displaystyle \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n}$ exist.

So we have $a_{n+1} - a_n = (n+2016)^{2014}$ ,

$\begin{aligned} \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} &=& \lim_{n\to\infty} \dfrac{ (n+2016)^{2014}}{(n+1)^{2015} - n^{2015}} \\ &=& \lim_{n\to\infty} \dfrac{ n^{2014} +\ldots }{2015n^{2014} + \ldots} \\ &=& \dfrac1{2015} \end{aligned}$

which is a finite number. Hence $\displaystyle \lim_{n\to\infty} \dfrac1{n^{2015}} \sum_{k=1}^n (k+2015)^{2014} = \lim_{n\to\infty} \dfrac{a_n}{b_n} = \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} = \dfrac1{2015}$ .

Notice that $\frac{1}{n^{s}}\sum_{k=1}^{n}k^{s-1}=\sum_{k=1}^{n}(\frac{k}{n})^{s-1}\frac{1}{n}$ is the right Riemann sum of the function $f(x)=x^{s-1}$ over the interval $[0,1]$ that corresponds to the partition $\{0, \frac{1}{n}, \frac{2}{n}, ... \frac{n}{n}\}.$ Therefore $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\lim_{n\to\infty}\frac{1}{n^{s}}\sum_{k=1}^{n}k^{s-1}=\int_{0}^{1}x^{s-1}dx=\frac{1}{s}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(1)$

On the other hand, if $s>r+1$ then $\lim_{n\to\infty}\frac{1}{n^{s}}\sum_{k=1}^{n}k^{r}=\lim_{n\to\infty}\frac{1}{n^{s-r-1}}\frac{1}{n^{r+1}}\sum_{k=1}^{n}k^{r}=\lim_{n\to\infty}\frac{1}{n^{s-r-1}}*\int_{0}^{1}x^{r}dx=0*\frac{1}{r+1}=0\:\:\:\:\:\:\:\:\:\:\:\:\:\:(2)$

Now using (1) and (2) we get $\lim_{n\to \infty}\frac{1}{n^{2015}} \sum_{k=1}^{n}(k+2015)^{2014}= \sum_{i=0}^{2014}\bigg(\binom{2014}{i}*2015^{i}*\lim_{n\to \infty}\frac{1}{n^{2015}} \sum_{k=1}^{n}k^{2014-i}\bigg)=\frac{1}{2015}.$

Therefore, the answer for this question is $1+2015=2016.$

Another interesting problem!

Just an outline of a solution: The sum will be a polynomial in $n+2015$ with leading term $\frac{(n+2015)^{2015}}{2015}$ , but that is just a polynomial in $n$ with leading term $\frac{n^{2015}}{2015}$ . After dividing by $n^{2015}$ we end up with $\frac{1}{2015}$ plus stuff that goes to 0 as $n\rightarrow\infty$ , so that the limit is $\frac{1}{2015}$ , with $a+b=\boxed{2016}$ .