Weird Limit

It is first important to notice that the limit in question is only well defined if $a > 1$ so we may ignore any kind of subtle problems that arise by including 0 or negative numbers and immediately assume $a>1$ . Therefore, we can focus our efforts on proving that for the limit to be true no other special property must be granted to $a$ .

Leaving x fixed and applying Euler's summation formula, $\sum_{i=1}^{a-1} i^x = 1 + \sum_{1 < i \leq a-1} i^x = 1 + \int_{1}^{a-1} i^x di + \int_{1}^{a-1} (t - \lfloor t \rfloor ) (i^x)' di = 1 + \frac{(a-1)^{x+1}}{x+1} - \frac{1}{x+1} + O((a-1)^x) = \frac{(a-1)^{x+1}}{x+1} + O((a-1)^x)$ .

First I will prove that the error term vanishes: $\lim_{x \to \infty} \frac{(a-1)^x}{a^x} = \lim_{x \to \infty} \left(\frac{a-1}{a} \right)^x = \lim_{x \to \infty} \left( 1 - \frac{1}{a} \right)^x = 0$ because as $a>1$ , the number inside the exponent is between 0 and 1 and thus vanishes as we increase $x$ .

Now, to prove that the main term vanishes, $\lim_{x \to \infty} \frac{ (a-1)^{x-1} }{(x+1) a^x }= \lim_{x \to \infty} \left(\frac{a-1}{a} \right)^{x-1} \frac{1}{a(x+1)}$ and this is a product of two limits that go to 0. Thus, assuming $a>1$ is enough.