Problem 7 IMC 2015 - Integral and Limit I'm trying to solve problem 7 from the IMC 2015, Blagoevgrad, Bulgaria (Day 2, July 30). Here is the problem

Compute
  $$\large\lim_{A\to\infty}\frac{1}{A}\int_1^A A^\frac{1}{x}\,\mathrm dx$$

And here is my approach:
Since $A^\frac{1}{x}=\exp\left(\frac{\ln A}{x}\right)$, then using Taylor series for exponential function, we have
$$\exp\left(\frac{\ln A}{x}\right)=1+\frac{\ln A}{x}+\frac{\ln^2 A}{x^2}+\frac{\ln^3 A}{x^3}+\cdots=1+\frac{\ln A}{x}+\sum_{k=2}^\infty\frac{\ln^k A}{x^k}$$
Hence, integrating term by term is trivial.
\begin{align}
\lim_{A\to\infty}\frac{1}{A}\int_1^A A^\frac{1}{x}\,\mathrm dx&=\lim_{A\to\infty}\frac{1}{A}\int_1^A \left[1+\frac{\ln A}{x}+\sum_{k=2}^\infty\frac{\ln^k A}{x^k}\right]\,\mathrm dx\\
&=\lim_{A\to\infty}\frac{1}{A}\left[A-1+\ln^2 A-\sum_{k=2}^\infty\frac{\ln^k A}{(k-1)A^{k-1}}+\sum_{k=2}^\infty\frac{\ln^k A}{(k-1)}\right]\\
&=1+\lim_{A\to\infty}\sum_{k=2}^\infty\frac{\ln^k A}{A(k-1)}\\
\end{align}
I'm stuck here because I can't evaluate the last term as $k\to\infty$. My guess the answer is $1$, but I'm not sure. Is my approach correct? If it's correct, how does one evaluate the last limit? I'm also interesting in knowing the other approaches of this this problem in formally math setting. Thanks.

For the integral in question:
$$\large\lim_{A\to\infty} \, \frac{1}{A} \, \int_{1}^{A} A^{\frac{1}{x}}\,dx$$
the following is considered:
\begin{align}
A^{\frac{1}{x}} = e^{\frac{1}{x} \, \ln A} = 1 + \frac{\ln A}{x} + \sum_{k=2}^{\infty} \frac{\ln^{k}A}{k! \, x^{k}}
\end{align}
for which 
\begin{align}
\int_{1}^{A} A^{\frac{1}{x}} \, dx &= [ x ]_{1}^{A} - \ln A \, [ \ln x ]_{1}^{A} + \sum_{k=2}^{\infty} \frac{\ln^{k}A}{k! \, (1-k)} \, [ x^{1-k} ]_{1}^{A} \\
&= A - 1 + \ln A - \sum_{k=2}^{\infty} \frac{1}{k! \, (k-1)} \, \left(\frac{A \, \ln^{k}A}{A^{k}} - \frac{\ln^{k}A}{1} \right).
\end{align}
Now dividing by $A$ leads to
\begin{align}
\frac{1}{A} \, \int_{1}^{A} A^{\frac{1}{x}} \, dx &= 1 - \frac{1}{A} + \frac{\ln A}{A} - \sum_{k=2}^{\infty} \frac{1}{k! \, (k-1)} \, \left(\frac{ \ln^{k}A}{A^{k}} - \frac{\ln^{k}A}{A} \right).
\end{align}
In order to evaluate the neccessary limits the following components are needed:
\begin{align}
\lim_{A \to \infty} \frac{\ln A}{A} &\to 0 \\
\lim_{A \to \infty} \frac{\ln^{k}A}{A^{k}} &= \left( \lim_{A \to \infty} \frac{\ln A}{A} \right)^{k} \to 0 \\
\lim_{A \to \infty} \frac{\ln^{k}A}{A} &= k \, \lim_{A \to \infty} \frac{\ln^{k-1}A}{A} = \cdots = \frac{k!}{1!} \, \lim_{A \to \infty} \frac{\ln A}{A} = 0.   
\end{align}
With these limiting values it is evident that
\begin{align}
\lim_{A \to \infty} \, \frac{1}{A} \, \int_{1}^{A} A^{\frac{1}{x}} \, dx &= 
\lim_{A \to \infty} \left[ 1 - \frac{1}{A} + \frac{\ln A}{A} - \sum_{k=2}^{\infty} \frac{1}{k! \, (k-1)} \, \left(\frac{ \ln^{k}A}{A^{k}} - \frac{\ln^{k}A}{A} \right) \right] \\
&= 1 - \lim_{A \to \infty} \frac{1}{A} - \sum_{k=2}^{\infty} \frac{1}{k! \, (k-1)} \, \lim_{A \to \infty} \left(\frac{ \ln^{k}A}{A^{k}} - \frac{\ln^{k}A}{A} \right) \\
&= 1.
\end{align}
