Nasty Limit of sum at infinity $$\lim_{n \to \infty} \left (\sum_{i=1}^n \frac{a\left[\left(\frac{b}{a}\right)^{\frac{i}{n}}-\left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) $$
How would I even begin to approach this??
Edit:
So the hints helped... I ended up with $\space\ln\left(\frac{b}{a}\right)$
 A: $$\begin{align}
&\lim_{n \to \infty} \left( \sum_{i=1}^n \frac{a \left[ \left(\frac{b}{a}\right)^{\frac{i}{n}} - \left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) \\ =
&\lim_{n \to \infty} \left( \sum_{i=1}^n \frac{ \left(\frac{b}{a}\right)^{\frac{i}{n}} - \left(\frac{b}{a}\right)^{\frac{i-1}{n}}}{\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) \\= &\lim_{n \to \infty} \left( \sum_{i=1}^n \frac{\left(\frac{b}{a}\right)^{\frac{i}{n}}}{\left(\frac{b}{a}\right)^{\frac{i-1}{n}}} - 1 \right) \\ =&\lim_{n \to \infty} \sum_{i=1}^n \left[ \left(\frac{b}{a}\right)^{\frac1n} - 1 \right] \\ =&\lim_{n \to \infty} n\left(\frac{b}{a}\right)^{\frac{1}{n}} - n \\ =&\lim_{n \to \infty} \frac{\left(\frac{b}{a}\right)^\frac{1}{n} - 1}{\frac{1}{n}} \\ =&\lim_{n \to \infty} \frac{\ln\left(\frac{b}{a}\right)\cdot \frac{-1}{n^2} \cdot \left(\frac{b}{a}\right)^\frac{1}{n}}{\frac{-1}{n^2}} \\= &\lim_{n \to \infty} \ln\left(\frac{b}{a}\right) \cdot \left(\frac{b}{a}\right)^{\frac{1}{n}} \\ =& \space\ln\left(\frac{b}{a}\right)
\end{align}$$
A: $$\lim_{n \to \infty} \left (\sum_{i=1}^n \frac{a\left[\left(\frac{b}{a}\right)^{\frac{i}{n}}-\left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right)  = \lim_{n \to \infty} \left (\sum_{i=1}^n \left(\left(\frac{b}{a}\right)^{\frac{1}{n}}-1\right)\right) = \lim_{n \to \infty} n \left(\frac{b}{a}\right)^{\frac{1}{n}}-n = \lim_{n \to \infty} \frac{\left(\frac{b}{a}\right)^{\frac{1}{n}}-1}{\frac{1}{n}} \overset{Hop.}{=} \lim_{n \to \infty}  \left(\frac{b}{a}\right)^{\frac{1}{n}}\text{ln}\left(\frac{b}{a}\right) = \text{ln}\left(\frac{b}{a}\right).$$
