$f_n:[0,\infty)\to\mathbb{R}$ is defined recursively by $f_1:=0$ and $$f_{n+1}(x)=e^{-2x}+\int_0^xf_n(t)e^{-2t}dt,\qquad n\ge 1$$ I need to show that the limit $f(x):=\lim_{n\to\infty} f_n(x)$ exists and find it explicitly.
I've found that $f_n$ is of the following form $$f_n(x)=\sum_{i=0}^{n-1} a_{n,i}e^{-2i}$$ and $$a_{n,i}=(-2)^{-i}\frac{a_{n-i+1,1}}{i!},\qquad \forall n,i\ge 2$$ But I can't find a function that looks like the limit.
The limit, if it exists, must satisfy $$f(x)=e^{-2x}+\int_0^xf(t)e^{-2t}dt$$ which after differentiation becomes $$f'(x) = -2e^{-2x} + f(x)e^{-2x}$$ This is a linear ODE you can solve. One approach is to write $g=f-2$, so that $g'(x)=g(x)e^{-2x}$; separate and integrate, eventually arriving at $$ f(x) = 2 +C \exp\left(-\frac12 \exp(-2x)\right) $$
The existence of limit follows from the proof of Picard uniqueness theorem for ODE, since the formula for $f_n$ is the Picard iteration.