Tight upper bound on $\sum_{j=1}^{n}(-1)^{j-1}{n \choose j}\frac{1}{j}$?

Question

I would like to prove the following inequality (upper bound):

\begin{align} \xi(n) = \sum_{j=1}^{n}(-1)^{j-1}{n \choose j}\frac{1}{j} \leq 3\ln(n), \end{align}

for $n\geq 2$. I have made an attempt to prove this using induction, i.e.

$(n=2):$ \begin{align} (-1)^{0}{2 \choose 1}\frac{1}{1} + (-1)^{1}{2 \choose 2}\frac{1}{2} = \frac{3}{2} \leq 3\ln(2). \end{align}

Assume $(n)$ holds and show $(n+1)$ holds: \begin{align} \xi(n+1) = \sum_{j=1}^{n+1}(-1)^{j-1}{n+1 \choose j}\frac{1}{j} &= (-1)^n\frac{1}{n+1} + \sum_{j=1}^{n}(-1)^{j-1}{n+1 \choose j}\frac{1}{j}\\ &= (-1)^n\frac{1}{n+1} + \sum_{j=1}^{n}(-1)^{j-1}\left[{n\choose j}+{n \choose j-1}\right]\frac{1}{j}\\ &= (-1)^n\frac{1}{n+1} + \sum_{j=1}^{n}(-1)^{j-1}{n\choose j}\frac{1}{j} + \sum_{j=1}^{n}(-1)^{j-1}{n \choose j-1}\frac{1}{j}.\\ &= (-1)^n\frac{1}{n+1} + \xi(n) + \sum_{j=1}^{n}(-1)^{j-1}{n \choose j-1}\frac{1}{j}.\\ \end{align}

Now I cannot figure out how \begin{align} (-1)^n\frac{1}{n+1} + \sum_{j=1}^{n}(-1)^{j-1}{n \choose j-1}\frac{1}{j} \leq 3\ln\left(1+\frac{1}{n}\right). \end{align}

Which is the missing part in this proof. Maybe I am on the wrong track. Any help is really appreciated.

Answer 1

I followed a totally different path. Hope it helps, anyway. $$\sum _{j=1}^n \frac{(-1)^{j-1} \binom{n}{j}}{j}=H_n=\psi (n+1)+\gamma$$

where $\gamma$ is Euler-Mascheroni constant and $\psi$ is digamma function

as $n\to\infty$ $$\psi (n+1)+\gamma=\ln n+\gamma +\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)\le 1+\ln n$$

As $\gamma +\frac{1}{2 n}<1.1$