Prove that $$ \sum_{r=1}^{2022}\frac{(-1)^{r-1}}{r}\binom{2022}{r} = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2022} $$
I have been trying to prove this by using the expansion of $(1-x)^{2022}$ after which the use of integral is what i was advised on by a few, so i wanted to know how it can be implemented here in the further continuation of the expansion
Consider $N\ge 0$ and define $\displaystyle F_N(x) = \sum_{r = 1}^{N}\frac{x^{r - 1}}{r}\binom{N}{r}$. Then your question concerns the value $F_{2022}(-1)$. We have $$ (F_N(x) x)' = \sum_{r = 1}^N x^{r - 1}\binom{N}{r} = \frac{(x + 1)^{N} - 1}{x}. $$ Notice that $$ \int \frac{(x + 1)^{n + 1} - 1}{x}\, dx - \int \frac{(x + 1)^{n} - 1}{x}\, dx = \int (x + 1)^ndx; \\ \int_{-1}^0 \frac{(x + 1)^{n + 1} - 1}{x}\, dx - \int_{-1}^0 \frac{(x + 1)^{n} - 1}{x}\, dx = \int_{-1}^0 (x + 1)^ndx = \frac{1}{n + 1}. $$ Inducting on the last line we get $$ \int_{-1}^0 \frac{(x + 1)^{n} - 1}{x}\, dx = 1 + \frac{1}{2} + \ldots + \frac{1}{n} = \sum_{r = 1}^n\frac{1}{r}. $$ Consequently $$ F_N(-1) = \int_{-1}^0(F_N(x)x)'dx = \int_{-1}^0 \frac{(x + 1)^{N} - 1}{x}\, dx = \sum_{r = 1}^N\frac{1}{r}. $$
