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.
 A: 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}
A: The key to showing that this limit converges to $1$ is to show that the function $A^{1/x}$ decreases quickly towards $1$ as $x$ increases beyond $1$, so that the integral isn't significantly larger than $\int_1^A 1\,dx \sim A$.  To this end we will focus on showing that the difference (which is clearly positive) is small enough:
$$\int_1^A (A^{1/x} - 1)\,dx = o(A).$$
This can be done be with three piecewise-constant approximations:
1) We first consider $x$ close to $1$, where we don't have an upper bound better than $A^{1/x} \le A$.  So to get a contribution of $o(A)$ we need to cut it off at $x = 1 + o(1)$.  We want $A^{1/x}$ to shrink down to a decent amount, so it behooves us to choose the $o(1)$ to be very slowly shrinking, say at $x = 1 + 2/\sqrt{\ln A}$. Then the contribution from this piece is
$$\int_1^{1 + 2/\sqrt{\ln A}} (A^{1/x} - 1)\, dx \le \int_1^{1 + 2/\sqrt{\ln A}} A\, dx = \frac{2A}{\sqrt{\ln A}} = o(A).$$
2) Now we consider $x \ge 1 + 2/\sqrt{\ln A}$.  For large enough $A$, we have $1/x \le 1 - 1/\sqrt{\ln A}$.  This means for $x \ge 1 + 2/\sqrt{\ln A}$,
$$\exp\big(\tfrac{\ln A}{x}\big) \le A \exp\big(-\tfrac{\ln A}{\sqrt{\ln A}}\big) = A \exp(-\sqrt{\ln A}).$$
This is considerably smaller than $A$, so we are free to use this bound for a sizable range of $x$.  We'd like to go beyond $x = \ln A$ which is where $A^{1/x}$ actually gets close to $1$.  Happily, this goal is in reach as (for all large $A$) $\sqrt{\ln A} > 3 \ln \ln A$, so
$$ \exp(-\sqrt{\ln A}) < \exp(-3 \ln \ln A) = 1/(\ln A)^3.$$
That means we can set the next cutoff at $x = (\ln A)^2$, and we still get
$$\int_{1 + 2/\sqrt{\ln A}}^{(\ln A)^2} A^{1/x} - 1 \, dx < \int_0^{(\ln A)^2} A \exp(-\sqrt{\ln A})\, dx < A \frac{(\ln A)^2}{(\ln A)^3} = o(A).$$
3) Now the last piece from $x = (\ln A)^2$ to $A$.  We chose the previous cutoff to make $A^{1/x}$ very close to one, indeed for $x \ge (\ln A)^2$ we have
$$A^{1/x} \le \exp(1/\ln A) = 1 + o(1),$$
since $\exp(1/\ln A) \to 1$ as $A \to \infty$ (we could be more precise using Taylor expansion and say that the RHS is $1 + O(1/\ln A)$).  Therefore the contribution from this third interval is
$$\int_{(\log A)^2}^A (A^{1/x} - 1)\, dx \le \int_0^A (\exp(1/\ln A) - 1)\, dx. = \int_0^A o(1)\, dx = o(A).$$
Combining 1), 2), 3) we have the desired upper bound.  Although the calculations are a little messy the argument is elementary and doesn't require integrating infinite series term-by-term.
