Consider the sum $$S_n=\sum_{r=2}^n \frac{(-1)^r\binom{n}{r}}{r-1} $$
My question is: What is the asymptotic for $S_n$ as $n\to \infty$ or find $R_n$ such that $S_n\sim R_n$ as $n\to \infty$.
My try:
Define the $n$th Harmonic number as $$H_n=\sum_{r=1}^n \frac{1}{r}$$ Then we also have (see https://en.m.wikipedia.org/wiki/Harmonic_number ) $$H_n=\sum_{r=1}^{n} \frac{(-1)^{r-1} \binom{n}{r} }{r} $$
Then we have the asymptotic equivalence of $H_n$ as $$H_n\sim \log n+\gamma +\mathcal{O}\left(\frac{1}{n}\right)$$
How to find the asymptotic equivalence of $S_n$?
Best Answer
Starting with the binomial identity $$(1-x)^n = \sum_{r = 0}^{n}(-1)^r\dbinom{n}{r}x^r,$$ we can subtract $1-nx$ from both sides and divide both sides by $x^2$ to get $$\dfrac{(1-x)^n-1+nx}{x^2} = \sum_{r = 2}^{n}(-1)^r\dbinom{n}{r}x^{r-2}.$$ Integrating both sides from $0$ to $1$ yields $$\int_{0}^{1}\dfrac{(1-x)^n-1+nx}{x^2}\,dx = \sum_{r = 2}^{n}\dfrac{(-1)^r\binom{n}{r}}{r-1} = S_n.$$ So we can find asymptotic for $S_n$ using the above integral. We split the integral into two pieces. One over $[0,1/n]$ and one over $[1/n,1]$.
For the first piece, one can check that $f(x) := \tfrac{(1-x)^n-1+nx}{x^2}$ is decreasing on $[0,1/n]$, and so $$\dfrac{1}{n}f\left(\dfrac{1}{n}\right) \le \int_{0}^{1/n}f(x)\,dx \le \dfrac{1}{n}f(0),$$ $$\dfrac{1}{n} \cdot n^2\left(1-\tfrac{1}{n}\right)^n \le \int_{0}^{1/n}\dfrac{(1-x)^n-1+nx}{x^2}\,dx \le \dfrac{1}{n} \cdot \dbinom{n}{2}$$ $$n\left(1-\tfrac{1}{n}\right)^n \le \int_{0}^{1/n}\dfrac{(1-x)^n-1+nx}{x^2}\,dx \le \dfrac{n-1}{2}$$ $$\dfrac{n-1}{e} \le \int_{0}^{1/n}\dfrac{(1-x)^n-1+nx}{x^2}\,dx \le \dfrac{n-1}{2},$$ where we have slightly weakened the lower bound from $n\left(1-\tfrac{1}{n}\right)^n$ to $e^{-1}(n-1)$ in the last step.
For the second piece, we note that $0 \le (1-x)^n \le \left(1-\tfrac{1}{n}\right)^n \le e^{-1}$ for all $x \in [1/n,1]$, and so $$\int_{1/n}^{1}\dfrac{-1+nx}{x^2}\,dx \le \int_{1/n}^{1}\dfrac{(1-x)^n-1+nx}{x^2}\,dx \le \int_{1/n}^{1}\dfrac{e^{-1}-1+nx}{x^2}\,dx$$ $$n\log n - (n-1) \le \int_{1/n}^{1}\dfrac{(1-x)^n-1+nx}{x^2}\,dx \le n\log n - \left(1-\dfrac{1}{e}\right)(n-1)$$
Combining these two bounds yields: $$n\log n - \left(1-\dfrac{1}{e}\right)(n-1) \le \int_{0}^{1}\dfrac{(1-x)^n-1+nx}{x^2}\,dx \le n\log n - \left(\dfrac{1}{2}-\dfrac{1}{e}\right)(n-1),$$ and thus, $S_n = n\log n + O(n)$.