For $a > 1$ determine the definite integral
$$ \int_1^a ln(x) $$
as the limit of a Riemann sum. Hint: Use the partitioning $P_N = (x_0, x_1, …, x_N)$ with:
$$ 1 = x_0 < x_1 < … < x_N = a \quad \text{and} \quad x_k = a^{\frac{k}{N}} \quad k = (0, 1, 2, …, N) $$
and make use of geometric and telescoping series.
My approach is to form a lower (or upper) Riemann sum and then taking the limit with $N \rightarrow \infty$. So far I have this:
$$ \begin{align*} S_{lower} &= \sum_{k=0}^{N-1} \inf f([x_k, x_{k+1}]) \cdot (x_{k+1} - x_k) \\ &= \sum_{k=0}^{N-1} f(x_k) \cdot (a^{\frac{k+1}{N}} - a^{\frac{k}{N}}) \\ &= \sum_{k=0}^{N-1} ln(a^{\frac{k}{N}}) \cdot (a^{\frac{k}{N} + \frac{1}{N}} - a^{\frac{k}{N}}) \\ &= \sum_{k=0}^{N-1} \frac{k}{N} \cdot ln(a) \cdot a^{\frac{k}{N}} (a^{\frac{1}{N}} - 1) \\ &= \frac{ln(a) (a^{\frac{1}{N}} - 1)}{N} \sum_{k=0}^{N-1} k \cdot (a^{\frac{1}{N}})^k \qquad \text{define } \quad q := a^{\frac{1}{N}} \\ &= \frac{ln(a) (q - 1)}{N} \sum_{k=0}^{N-1} k \cdot q^k \qquad \text{apply formula for partial sum of geometric series} \\ &= \frac{ln(a) (q - 1)}{N} \sum_{k=0}^{N-1} q \cdot \frac{d}{dq} q^k \\ &= \frac{ln(a) (q - 1)}{N} \cdot q \cdot \frac{d}{dq} \sum_{k=0}^{N-1} q^k \\ &= \frac{ln(a) (q - 1)}{N} \cdot q \cdot \frac{d}{dq} \frac{1 - q^N}{1 - q} \\ &= \frac{ln(a) (q - 1)}{N} \cdot \frac{(N+1)q^{N+1} - Nq^N - q}{(1-q)^2} \qquad \text{resubstitute q} \\ &= \frac{ln(a) (a^{\frac{1}{N}} - 1)}{N} \cdot \frac{(N+1)(a^{\frac{1}{N}}) ^{N+1} - N (a^{\frac{1}{N}})^N - a^{\frac{1}{N}}}{(1-a^{\frac{1}{N}})^2} \end{align*} $$
Not really sure where to go from there. Obviously if I take the limit as $N \rightarrow \infty$ now I would have some zeroes in bad places. I believe I‘m starting off right but taking a wrong turn somewhere. Any help would be greatly appreciated!