How do I prove that the -1 constant can be ignored in this summation for a limit?

Question

I effectively want to show that:

$$\lim_{n \to \infty} \sum_{i=an}^{n} \frac{1}{i-1} = \lim_{n \to \infty} \sum_{i=an}^{n} \frac{1}{i}$$

I know that $$\frac{1}{n-1} < \frac{1}{i-1} < \frac{1}{an-1}$$ Also $0 < a < 1$.

I first split the first summation up into two separate sums like so:

$$\sum_{i=an}^{n} \frac{1}{i-1} = \sum_{i=an}^{n} \frac{1}{i} + \sum_{i=an}^{n} \frac{1}{i^2}$$

So I guess I have to show the second summation $\to$ 0 as $n \to \infty$?

Answer 1

$$\lim_{n \to \infty} \left\{\sum_{i\ge an}^{n} \frac{1}{i-1} - \sum_{i\ge an}^{n} \frac{1}{i}\right\}=\frac1{\lceil an-1\rceil}-\frac1{n}$$ This difference goes to $0$.

Answer 2

Hint: $$\sum_{i=n}^{m} \frac{1}{i-1} = \sum_{i=n-1}^{m-1} \frac{1}{i}$$ So you gain one term at one end and lose one at the other.