Let $f(n)$ be a function defined for $n\ge 2$ and $n\in N$ which follows the recurrence(for $n\ge 3$) $$\displaystyle f(n)=f(n-1) +\displaystyle \frac {4\cdot (-1)^{(n-1)} \cdot \left(\displaystyle \sum_{d \vert (n-1)} (\chi (d))\right) }{n-1}$$ where $d\vert (n-1)$ means $d$ divides $(n-1)$ i.e. $d$ is divisor of $(n-1)$ .Also assume that $f(2)=-4$.
Where I define $$\chi(d) = \begin{cases} 1, & \text{if $d=4k+1$ where $k$ is a whole number} \\ -1, & \text{if $d=4k+3$ where $k$ is a whole number} \\ 0, & \text {if $d$ is even natural number} \end{cases}$$. Then find $$\lim_{n\to \infty} f(n)$$
First of all this is not at all an assignment or homework problem. It is just a question I came up with, when I was playing with a limit consisting of tedious geometry.
Second thing, I tried to find an explicit formula for the function but it seems impossible for me. Also I tried to use the recurrence and guess the approaching value. But the function I guess approaches to some limit (which I don't know) very slowly and hence I am not able to guess the limit.
Any guidance and help towards the solution would be quite helpful.
A preliminary lemma relates your $\chi$ function with the Gaussian integers:
$$ 4\sum_{d\mid n}\chi(d) = r_2(n) = \left|\{(a,b)\in\mathbb{Z}^2:a^2+b^2=n\}\right| $$ hence your question is equivalent to the determination of the Dirichlet L-series $$ L=\sum_{n\geq 1}\frac{(-1)^{n+1} r_2(n)}{n} $$ which is conditionally convergent convergent by Gauss circle problem: the average value of $r_2(n)$ is $\pi$, i.e. the area of the unit circle. Since $r_2(2n)=r_2(n)$, the algebra of the Dirichlet series ensures that the wanted limit is given by
$$4\sum_{n\geq 1}\frac{1}{n}\sum_{d\mid n}\chi_4(d)\chi_2\left(\frac{n}{d}\right)=4 L(\chi_4,1)L(\chi_2,1)=\color{red}{-\pi\log 2} $$ where $\chi_4=\chi$ is the non-principal character $\!\!\!\pmod{4}$ and $\chi_2$ is the non-principal character $\!\!\!\pmod{2}$.