For $a> 0$, determine $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)$? For $a> 0$, determine  $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)$
My attempts  : $\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1) = ( a^\frac{1}{n^2}- 1) + ( a^\frac{2}{n^2}- 1)+......... +(a^\frac{n}{n^2}- 1)=a^\frac{1}{n^2}+ a^\frac{2}{n^2}++.....+a^\frac{n}{n^2}- n $
ThereFore  $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)= a^{\lim_{n\rightarrow \infty}\frac{1}{n^2}}+ a^{\lim_{n\rightarrow \infty}\frac{2}{n^2}}++.....+a^{\lim_{n\rightarrow \infty}\frac{n}{n^2}} - \lim_{n\rightarrow \infty} n = \lim_{n\rightarrow \infty} (a^0 + a^0 +.....+a^0) -\lim_{n\rightarrow \infty} n = \lim_{n\rightarrow \infty}( 1+1.....+1) -\lim_{n\rightarrow \infty} n=\lim_{n\rightarrow \infty} n -\lim_{n\rightarrow \infty} n = 0 $
Hence $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}( a^\frac{k}{n^2} - 1)= 0$
Is  my answer  is correct  ??
Any  hints / solution will be appreciated
thanks u..
 A: Intuition:
since $e^u = 1+u + o(u)$ when $u\to 0$ and 
$0 \leq \frac{k}{n^2}\ln a \leq \frac{\ln a}{n} \xrightarrow[n\to\infty]{} 0$ for every $1\leq k\leq n$, we "expect" to have
$$
\sum_{k=1}^{n}( a^{\frac{k}{n^2}}-1)
= \sum_{k=1}^{n}( e^{\frac{k}{n^2}\ln a}-1) \approx  \sum_{k=1}^{n} \frac{k}{n^2}\ln a = \ln a\cdot \frac{n(n+1)}{2n^2}\xrightarrow[n\to\infty]{} \frac{\ln a}{2} \tag{1}
$$
so we should try and make this rigorous. (Sanity check: this is consistent with $a=1$, for which we do get $0$.)
Side note. Easy lower bound: using the standard inequality (proven e.g. by convexity) $e^x \geq 1+x$ for all $x$, we have
$$
\sum_{k=1}^{n}( e^{\frac{k}{n^2}\ln a}-1) \geq \sum_{k=1}^{n}\frac{k}{n^2}\ln a = \frac{\ln a}{2}\cdot \frac{n(n+1)}{n^2} \tag{2}
$$
and as mentioned we have $\lim_{n\to\infty} \frac{\ln a}{2}\cdot \frac{n(n+1)}{n^2} = \frac{\ln a}{2}$. So we have that the limit is at least $\frac{\ln a}{2}$... That was the easy part.
Proof of (1).
We can write
$$\begin{align}
\sum_{k=1}^{n}( e^{\frac{k}{n^2}\ln a}-1) 
&=\sum_{k=1}^{n}{\underbrace{\big(e^{\frac{\ln a}{n^2}}\big) }_{=\alpha}}^k - n = \alpha\cdot \frac{1-\alpha^{n}}{1-\alpha} - n
= e^{\frac{\ln a}{n^2}}\frac{1-e^{\frac{\ln a}{n}}}{1-e^{\frac{\ln a}{n^2}}} - n \\
&= \left(1+ o\left(\frac{1}{n}\right)\right)\cdot \frac{\frac{\ln a}{n}+\frac{\ln^2 a}{2n^2}+o(\frac{1}{n^2})}{\frac{\ln a}{n^2}+o(\frac{1}{n^2})} - n\\
&= \left(1+ o\left(\frac{1}{n}\right)\right)\cdot \frac{n+\frac{\ln a}{2}+o(1)}{1+o(1)} - n\\
&= n+\frac{\ln a}{2}+o(1) - n\\
&= \frac{\ln a}{2}+o(1) \xrightarrow[n\to\infty]{} \frac{\ln a}{2} \tag{3}
\end{align}$$
where we used the Taylor series of $e^u$ as $u\to 0$, $e^u = 1+u+\frac{u^2}{2}+o(u)$, and that of $\frac{1}{1+u} = 1-u+o(u)$. This last (3) gives us the expected limit, $\frac{\ln a}{2}$.
