I am looking at the proof that, if $S_0,S_1,\ldots,S_n$ are the partial sums of a random walk, then: $$\mathbb{P}(\Lambda) = \mathbb{P}(S_0,\ldots, S_n\text{ has a point of increase})\leq\frac{C}{\log(n)}$$
The proof of the theorem goes as follows: it is known that $\mathbb{P}(\Lambda)\leq 2\frac{\sum_{k=0}^{n}p_kp_{n-k}}{\sum_{k=0}^{\underline{n/2}}p_k^2}$, where $\underline{a}$ is the floor function of $a$, and where $p_k = \mathbb{P}(S_i\geq 0\text{ for all }1\leq i\leq k)$. Given this, we can upper bound the numerator by the following: $$\sum_{k=0}^{n}p_kp_{n-k}\leq 2 + 4C_2^2n^{-1/2}\sum_{k=1}^{\underline{n/2}}k^{-1/2}$$ For some positive constant $C_2$. We can then recognize that $p_k^2\geq \frac{C_1^2}{n}$ for another positive constant $C_1$, and hence, that $\sum_{k=0}^{\underline{n/2}}p_k^2 \geq C_1^2\log(\underline{n/2})$, since $\sum_{k=1}^{n}1/k\leq \log(n)$. So we now have the following bound: $$\mathbb{P}(\Lambda)\leq 2\frac{2 + 4C_2^2n^{-1/2}\sum_{k=1}^{\underline{n/2}}k^{-1/2}}{C_1^2\log(\underline{n/2})}$$ How do I get from this bound to $C/\log(n)$? The book that I am reading does not seem to justify this jump, but claims that the above expression completes the proof, which I do not find self-evident.
Note that $\sum_{k=1}^{n/2} k^{-1/2}$ is an upper Riemann sum for the integral $$\int_1^{n/2} x^{-1/2}dx = 2(n/2)^{-1/2} - 2$$ So we get $$\mathbb{P}(\Lambda) \leq 2 \frac{2 + C_3 - C_4n^{-1/2}}{C_1^2\log(n) - C_1^2\log(2)} \leq 2 \frac{2 + C_3}{C_1^2\log(n) - C_1^2\log(2)} $$ At this point, we can find a constant $C$ such that for $n \geq N$, $$\mathbb{P}(\Lambda) \leq 2 \frac{2 + C_3}{C_1^2\log(n) - C_1^2\log(2)} \leq C/\log(n).$$ Then, make the constant $C$ work for the finitely many $n < N.$