Nasty límite de suma al infinito

Question

$$\lim{n \to \infty} \left (\sum{i=1}^n \frac{a\left[\left(\frac{b}{a}\right)^{\frac{i}{n}}-\left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) $$

¿¿Incluso empiezo a acercarse a esto??

Edit: Así los consejos ayudó a... Terminé con $\space\ln\left(\frac{b}{a}\right)$

Answer 1

$$\begin{align} &\lim{n \to \infty} \left( \sum{i=1}^n \frac{a \left[ \left(\frac{b}{a}\right)^{\frac{i}{n}} - \left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) \ = &\lim{n \to \infty} \left( \sum{i=1}^n \frac{ \left(\frac{b}{a}\right)^{\frac{i}{n}} - \left(\frac{b}{a}\right)^{\frac{i-1}{n}}}{\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) \= &\lim{n \to \infty} \left( \sum{i=1}^n \frac{\left(\frac{b}{a}\right)^{\frac{i}{n}}}{\left(\frac{b}{a}\right)^{\frac{i-1}{n}}} - 1 \right) \ =&\lim{n \to \infty} \sum{i=1}^n \left[ \left(\frac{b}{a}\right)^{\frac1n} - 1 \right] \ =&\lim{n \to \infty} n\left(\frac{b}{a}\right)^{\frac{1}{n}} - n \ =&\lim{n \to \infty} \frac{\left(\frac{b}{a}\right)^\frac{1}{n} - 1}{\frac{1}{n}} \ =&\lim{n \to \infty} \frac{\ln\left(\frac{b}{a}\right)\cdot \frac{-1}{n^2} \cdot \left(\frac{b}{a}\right)^\frac{1}{n}}{\frac{-1}{n^2}} \= &\lim{n \to \infty} \ln\left(\frac{b}{a}\right) \cdot \left(\frac{b}{a}\right)^{\frac{1}{n}} \ =& \space\ln\left(\frac{b}{a}\right) \end {Alinee el} $$

Answer 2

$$\lim{n \to \infty} \left (\sum{i=1}^n \frac{a\left[\left(\frac{b}{a}\right)^{\frac{i}{n}}-\left(\frac{b}{a}\right)^{\frac{i-1}{n}}\right]}{a\left(\frac{b}{a}\right)^{\frac{i-1}{n}}}\right) = \lim{n \to \infty} \left (\sum{i=1}^n \left(\left(\frac{b}{a}\right)^{\frac{1}{n}}-1\right)\right) = \lim{n \to \infty} n \left(\frac{b}{a}\right)^{\frac{1}{n}}-n = \lim{n \to \infty} \frac{\left(\frac{b}{a}\right)^{\frac{1}{n}}-1}{\frac{1}{n}} \overset{Hop.}{=} \lim_{n \to \infty} \left(\frac{b}{a}\right)^{\frac{1}{n}}\text{ln}\left(\frac{b}{a}\right) = \text{ln}\left(\frac{b}{a}\right).$$