Question about $f(x)=\sum_{k=1}^\infty (-1)^{k+1}\sin (\frac{x}{k}) $

We could rewrite the function as $$ \sum_{k=1}^{\infty} \big(\sin\big(\frac{x}{2k-1}\big) - \sin\big(\frac{x}{2k}\big)\big) = 2 \sum_{k=1}^{\infty} \sin\big(\frac{x}{4k(2k-1)}\big) \cos\big(\frac{x(4k -1)}{4k(2k-1)}\big) $$

In the last term we find $k$ such that the inequality $$ \frac{x}{4k(2k-1)} \leq 1 $$

holds. It is achieved for $$ k > K^{\prime} = \frac{1 + \sqrt{2x + 1}}{4}. $$ First $K^{\prime}$ elements in the sum above could be bounded by $1$, and for the remainder of that sum we use the inequality $\sin{y} \leq y$. We get $$ 2 \sum_{k=1}^{\infty} \sin\big(\frac{x}{4k(2k-1)}\big) \cos\big(\frac{x(4k -1)}{4k(2k-1)}\big) \leq $$ $$ 2K^{\prime} + 2\sum_{k > K^{\prime}} \frac{x}{4k(2k-1)} $$ $$ \leq 2K^{\prime} + 2 \frac{x}{4K^{\prime}(2K^{\prime}-1)} \sum_{k > K^{\prime}} \frac{1}{\frac{k}{K^{\prime}}\frac{2k-1}{2K^{\prime}-1}} $$ $$ \leq 2K^{\prime} + 2 \sum_{k > K^{\prime}} \frac{1}{\frac{k}{K^{\prime}}\frac{2k-1}{2K^{\prime}-1}}. $$

The last sum is the convergent series, it is bounded by $2K^{\prime}$, and it can be made very small for large $k$ and fixed $K^{\prime}$. So the rough estimate of the fuction is $O\big(\sqrt{x}\big)$, and we have to take more than $2K^{\prime}$ terms to get a good approximation.