How to prove that this integral operator converges to e^-x? I have this integral operator:
$ L[y]=\int_0^x (s-y(s) ds) $
Applying the operator to $y_0(x)=0$, I've found
$y_1(x)=(1/2)x^2$,
$y_2(x)=(1/2)x^2-(1/6)x^3$
$y_3(x)=(1/2)x^2-(1/4)x^3+(1/24)x^4$
It seems that $\lim\limits_{n\to\infty}   y_n(x)=1/ e^x$
But, given taylor series,  how can I prove this?
Thanks for your time
 A: Step 1: Finding the suspect
So, you are looking at
$$
y_0(x)=0,\quad \text{and}\quad y_n(x)=\int_0^x (s-y_{n-1}(s))\,ds.\tag{*}
$$
First we note that if $y_n$ converges to some function $y$, then so does $y_{n-1}$, and we could suspect that $y$ satisfies
$$
y(x)=\int_0^x(s-y(s))\,ds.
$$
Differentiating, we find that
$$
y'(x)=x-y(x),\quad y(0)=0.
$$
This is a linear differential equation, which has solution
$$
y(x)=e^{-x}-1+x.
$$
Thus, this function is clearly our candidate. 
Step 2: Prove it is guilty:
By doing a Taylor expansion of our suspect, we find that
$$
y(x)=\sum_{k=2}^{+\infty}\frac{(-1)^k}{k!}x^k.
$$
Now, what if we can show that the partial sums, up to a renumbering of one, coincide with the $y_n$? Let us try to show by induction that the $y_n$ defined in $(*)$ satisfies
$$
y_n(x)=\sum_{k=2}^{n+1}\frac{(-1)^k}{k!}x^k,\quad \forall n\geq 1.\tag{**}
$$
Now $y_1(x)=x^2/2$, so the first step is OK. Assume that it is true for $n-1$, i.e. that
$$
y_{n-1}(x)=\sum_{k=2}^{n}\frac{(-1)^k}{k!}x^k.
$$
Then, integrating, we find that (we integrate, and at one step put $l=k+1$)
$$
\begin{split}
y_n(x)&=\int_0^x(s-y_{n-1}(s))\,ds\\
&=\int_0^x \Bigl(s-\sum_{k=2}^{n}\frac{(-1)^k}{k!}s^k\Bigr)\,ds\\
&=\frac{x^2}{2}-\sum_{k=2}^n\frac{(-1)^k}{k!}\frac{x^{k+1}}{k+1}\\
&=\frac{x^2}{2}+\sum_{l=3}^{n+1}\frac{(-1)^l}{l!}x^l\\
&=\sum_{l=2}^{n+1}\frac{(-1)^l}{l!}x^l
\end{split}
$$
Thus, $(**)$ is proven by induction. Now, recognizing the series expansion of $e^{-x}$, we find that, indeed,
$$
y_n(x)\to e^{-x}-1+x
$$
as $n\to+\infty$.
