Evaluating: $\lim_{n\to \infty} \sum_{i=1}^n \frac{1}{\frac{an}{b-a}+i}$

Question

$$\lim_{n\to \infty} \sum_{i=1}^n \frac{1}{\frac{an}{b-a}+i},\ a,b\in\Bbb R\setminus\{0\},\ a\ne b$$

I know for a fact that the solution can be found via Laurent Series if that hint helps. I inserted a sample of the above with $a=1$ and $b=2$ on Wolfram and concluded that the result, as expected, is:

$$\ln\left|\frac{b}{a}\right|$$

The overall issue here is how to end up to the very above result!

Note: I'm a high senior school student.

Thank you!

Answer 1

Use a Riemann sum, i.e. note that equal-width strips gives $\int_0^1f(x)dx=\lim_{n\to\infty}\frac1n\sum_{i=1}^nf(i/n)$ so$$\begin{align}\lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{\frac{an}{b-a}+i}&=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{\frac{a}{b-a}+\frac{i}{n}}\\&=\int_{0}^{1}\frac{dx}{\frac{a}{b-a}+x}\\&=\left[\ln\left|\frac{a}{b-a}+x\right|\right]_{0}^{1}\\&=\ln\left|\frac{b}{b-a}\right|-\ln\left|\frac{a}{b-a}\right|\\&=\ln\left|\frac{b}{a}\right|.\end{align}$$

Answer 2

Before you changed your question you were asking to evaluate

\begin{equation}\label{Q} \lim_{n\to \infty} \sum_{i=1}^n \frac{1}{\frac{an}{b-a}+i} \tag{Q} \end{equation}

which is not the same as the series $\sum_{n=1}^\infty \frac{1}{\frac{an}{b-a}+i}$. The latter one clearly diverges by confrontation with $\sum_n \frac{1}{n}$.

The short answer for \ref{Q} above is that it is a riemann sum approximation of the integral $\int_{a}^{b} \frac{1}{x} dx$, but i guess you are willing to solve this only by elmentary methods.

Assume for simplicity $0<a<b$, then as you can check

\begin{equation}\label{in} \frac{b-a}{b} = 1-\frac{a}{b} < \ln(b/a) < \frac{b}{a}-1 = \frac{b-a}{a} \tag{1} \end{equation}

Now let $\displaystyle{x_i= a+(b-a)\frac{i}{n}}$. Note that $x_{i+1}-x_{i}=(b-a)/n$ and that the terms of the sum in your limit can be written as $$ \frac{1}{\frac{an}{b-a}+i} = \frac{b-a}{n} \frac{1}{a\frac{n-i}{n}+b\frac{i}{n}} = \frac{x_{i}-{x_{i-1}}}{x_i}$$

Now you can write, as follows from \ref{in},

$$\ln(x_{i+1}/x_{i}) <\frac{x_{i+1}-x_{i}}{x_{i}} = \frac{x_{i}-{x_{i-1}}}{x_i} < \ln(x_i/x_{i-1})$$

and get the following bound on the sums you are trying to get the limit of

$$\sum_{i=1}^n \frac{x_{i}-x_{i-1}}{x_{i}} < \sum_{i=1}^n \ln (x_{i}/x_{i-1}) = \ln(b/a)$$ $$\sum_{i=1}^n \frac{x_{i}-x_{i-1}}{x_{i}} > \sum_{i=1}^{n} \ln (x_{i+1}/x_{i}) = \ln(x_{n+1}/x_{1})=\ln\left(\frac{b+(b-a)/n}{{a+(b-a)/n}}\right)$$

At this point you only need to point out that $\ln\left(\frac{b+(b-a)/n}{{a+(b-a)/n}}\right) \rightarrow \ln(b/a)$ as $n\rightarrow \infty$, and use the "Squeeze theorem".

You can deduce what happens in the other cases from this. The cases in which $ab<0$ require some care.