Evaluate a limit with squeeze theorem

Question

Let $n\in\mathbb{N}, n\geq 2$ and $m\in[\![1;n-1]\!]$

Supposing that the $\frac{n}{m}$ stay constant and equal to $\lambda$, find the limit as $n \to +\infty$ of : $$ p_n(m) =\sum_{k=m}^{n-1} \frac{1}{k}$$

Using that for all $k\in [\![2;+\infty[\![$ : $\ln(k+1) - \ln(k) \leq \frac{1}{k} \leq \ln(k) - \ln(k-1)$, I have :

$$ \ln(\lambda) \leq p_n(m) \leq \ln\left(\frac{n-1}{m-1}\right)$$

I don't know how to continue, but I suppose that I have to show that $\lim\limits_{n \to +\infty} \ln\left(\frac{n-1}{m-1}\right) = \ln(\lambda)$, but I have some difficulties to show that.

Answer 1

Consider:

$$\ln \left (\frac{n-1}{m-1}\right )= \ln \left (\frac{\lambda m-1}{m-1}\right )=\ln \left (\frac{\lambda -1/m}{1-1/m}\right ).$$

As $n\to \infty,$ $m\to \infty,$ so the last expression $\to \ln \lambda.$