So I was trying to put $$\sum_{n=2}^\infty\frac{(-1)^n}{\pi(n)}$$(where $\pi(n)$ is the prime counting function) into Wolfram Alpha. It didn't work. So I decided to replace $\pi(n)$ with $\ln n$. I got this. But what's "i" doing here? (I couldn't add LaTeX or the post will look weird). I have no idea why Wolfram Alpha did this. Surely ln n will never be imaginary in the realm of numbers greater than 2, am I right?
Error estimates for these types of sums are often obtained by relating the sum to an integral which is then evaluated numerically. That introduces numerical error which can produce a very small imaginary portion.
You can enter the prime counting function $\pi(n)$ into WolframAlpha as
Thus, the following should estimate your desired sum:
sum (-1)^n/primepi(n), n=2..infI guess it's pretty easy to see that the answer should be 1 but we again see the numerical error.