Similar to this question Turning infinite sum into integral, I'm trying to express the following sum as an integral (as $n \to\infty $).
$\frac{r-1}{n} \sum^{n}_{i = r}\frac{1}{i-1}$.
I know the answer should be something close to $x\,\int_{x}^{1} 1/t \,dt$, but I'm not sure how the authors obtained this result.
You asked the question to write the following sum in the limit $n\to\infty$ as an integral, $$ \lim_{n\to\infty}\frac{r-1}{n}\cdot\sum_{i=r}^n\frac{1}{i-1}. $$ I understand that the lower summation bound $r=a n$ depends on $n$. The factor in front of the sum converges to $(r-1)/n \stackrel{n\to\infty}{\rightarrow}a$. The sum becomes $$ \lim_{n\to\infty}\sum_{i=an}^n\frac{1}{i-1} =\lim_{n\to\infty}\sum_{i=an}^{n-1}\frac{1}{i} $$ in the limit. Let me write the sum under the limit on the right-hand side as a Riemann sum for the definite integral $$ \int_{a/(1-a)^2}^{1/(1-a)^2} 1/t\,dt $$ with $n(1-a)$ subdivisions of constant widths $(n(1-a)^2)^{-1}$. Note that the cumulative width of the subdivisions is $(1-a)^{-1}$ as it should be. We have $$ \sum_{i=an}^{n-1}\frac{1}{i} =\sum_{i=an}^{n-1}\frac{1}{n(1-a)^2}\cdot\frac{1}{\frac{i}{n(1-a)^2}} =\sum_{i=0}^{n(1-a)-1}\frac{1}{n(1-a)^2}\cdot\frac{1}{\frac{i+na}{n(1-a)^2}}. $$ I rewrite the right-hand side into $$ \sum_{i=an}^{n-1}\frac{1}{i} =\sum_{i=0}^{n(1-a)-1}\frac{1}{n(1-a)^2}\cdot\frac{1}{\frac{a}{(1-a)^2}+\frac{i}{n(1-a)^2}}. $$ Using the number of subdivision $m:=n(1-a)$, we obtain $$ \sum_{i=an}^{n-1}\frac{1}{i} =\sum_{i=0}^{m-1}\frac{1}{m(1-a)}\cdot\frac{1}{\frac{a}{(1-a)^2}+\frac{i}{m(1-a)}}, $$ which converges to the integral suggested above, $$ \lim_{m\to\infty} \sum_{i=0}^{m-1}\frac{1}{m(1-a)}\cdot\frac{1}{\frac{a}{(1-a)^2}+\frac{i}{m(1-a)}} =\int_{a/(1-a)^2}^{1/(1-a)^2} 1/t\,dt = -\ln(a)\,. $$ The full expression above then becomes $$ \lim_{n\to\infty}\frac{r-1}{n}\cdot\sum_{i=r}^n\frac{1}{i-1} =-a\ln(a)\,. $$ It is now easy to check that $a=1/e$ maximizes this expression.
