Limit of series equals improper integral - IMC 2001

Question

I would like to know what result is used to justify the second to last inequality here (where the limit becomes the improper integral). It is the official solution from IMC 2001 (https://www.imc-math.org.uk/?year=2001&item=problems)

Answer 1

Imagine computing the area under the graph of $y=\frac{1}{1+e^x}$. If we dissect the region under the curve lying in the first quadrant into rectangles of width $h$ using vertical lines and approximate the height the height of each rectangle to be the value of the function $\frac{1}{1+e^x}$ at the right end of the rectangle, we can approximate the area under the curve as $$\sum_{n=1}^{\infty} \frac{h}{1+e^{nh}}$$ Approximation gets better and better as $h\to 0$, after all that is how we compute the area under the curve. Hence $$\lim_{h \to 0}\sum_{n=1}^{\infty} \frac{h}{1+e^{nh}} =\int_{0}^{\infty} \frac{dx}{1+e^x}=\ln 2$$

I remember solving this problem sometime ago, I even posted a solution but it got deleted for unjustifiable reasons. Anyway if you want to have a look at my solution, here it is:

first we calculate $P=\sum_{n=1}^{\infty} \frac{x^n}{1+x^n}$ for $x<1$, it is $$ \sum_{n=1}^{\infty} x^n(1+x^n)^{-1}=\sum_{n=1}^{\infty} x^n(1-x^n+x^{2n}+\cdots)=\sum_{n=1}^{\infty} \sum_{i=1}^{\infty} (-1)^{i+1} x^{in}=\sum_{i=1}^{\infty}(-1)^{i+1} \cdot \sum_{n=1}^{\infty} x^{in}$$ but $\sum_{n=1}^{\infty} x^{in}=\frac{x^i}{1-x^i}$, hence $$P=\sum_{i=1}^{\infty} \frac{(-1)^{i+1}x^i}{1-x^{i}} \implies (1-x)\cdot P=\sum_{i=1}^{\infty} \frac{(-1)^{i+1}x^i}{1+x+\cdots +x^i}$$ So $$\lim_{x \to 1^{-}} (1-x)P =\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}$$ Note that for $Q=\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}$, $$Q= \sum_{i=1}^{\infty} \int_{-1}^{0} x^{i-1}dx=\int_{-1}^{0} \left(\sum_{i=1}^{\infty} x^{i-1} \right)dx=\int_{-1}^{0} \frac{dx}{1-x}=\ln 2$$ So the final answer is $\ln 2$.