I know it's used to give values to divergent series, but when applied to absolutely convergent series, does it give the value at which the series converges?
By the way since someone asked about how I define the Ramanujan summation formula here it is: $\sum_{n=1}^{\infty} f(n)=-\frac{f(0)}{2}+i \int_{0}^{\infty} \frac{f(i t)-f(-i t)}{e^{2π t}-1} d t$
Yes. Ramanujan summation basically is the indefnite sum, $\sum_{n}f(n)=F(n)$ with the indefinite sum being true in the neighbourhood of f(n) which makes the solution unique, and $\sum_{n=a}^{b}f(n)=F(b)-F(a-1)$ We define the ramanujan sum value as a=1 so that $\sum_{n=a}^{b}f(n)=F(b)-F(0)$, if a sum is convergent then F(b) goes to infinity, and is 0. If the sum is divergent, then F(b) diverges and will be "disregarded", the constant associate value with the sum is then -F(0). Because F(0) is often ill defined, we can write it as, $-F(0)=-F(1)+f(1)$.
It would be super pointless as summation methode if this wasn't true. People make this topic so much harder then it needs to be.