The limit ofthatsequence is $\displaystyle e^x$ as you should be able to recognize the MacLaurin series for $\displaystyle e^x$

However, the derivative of $\displaystyle y= e^x$ is $\displaystyle e^x= y$, not x+ y.

If you are trying to use Picard's method to solve this equation, as Ackbeet suggested, then you are doing it wrong.

From dy/dx= x+y we get dy= (x+y)dx so that $\displaystyle y= \int (x+ y)dx$. Of course, since we don't know y but we can approximate it. If we are given y(0)= 1 (which you don't say but is implied by "y0= 1") as a first approximation, assume y(x)= 1 for all x. Then dy/dx= x+1 so that dy= (x+1)dx and [tex]y= (1/2)x^2+ x+ C[tex] and y(0)= C= 1 gives $\displaystyle y1= (1/2)x^2+ x+ 1$, as you have.

But then, $\displaystyle y2= \int x+ y1 dx= \int x+ (1/2)x^2+ x+ 1 dx= \int (1/2)x^2+ 2x+ 1 dx= (1/6)x^3+ x^2+ x+ C$ and, again, $\displaystyle y(0)= C= 1$ gives $\displaystyle y2= (1/6)x^2+ x^2+ x+ 1$ which is NOT what you have. I expect you forgot the added x.

Now, $\displaystyle y3= \int x+ y2 dx= \int x+ (1/6)x^3+ x^2+ x+ 1 dx$$\displaystyle = \int (1/6)x^3+ x^2+ 2x+ 1 dx= (1/24)x^4+ (1/3)x^3+ x^2+ x+ C$ and $\displaystyle y(0)= C= 1$ gives $\displaystyle y3= (1/24)x^4+ (1/3)x^3+ x^2+ x+ 1$.

One more time: $\displaystyle y4= \int x+ y3 dy= \int x+(1/24)x^3+ (1/3)x^3+ x^2+ x+ 1 dx$$\displaystyle = \int (1/24)x^4+ (1/3)x^3+ x^2+ 2x+ 1 dx= (1/5!)x^5+ (1/12)x^4+ (1/3)x^3+ x^2+ x+ C$ and $\displaystyle y(0)= C= 1$ gives

$\displaystyle y4= (1/5!)x^5+ (2/4!)x^4+ (2/3!)x^3+ (2/2!)x^2+ x+ 1$.

I think I would conjecture now that

$\displaystyle yn= (1/n!)x^n+ (2/(n-1)!)x^{n-1}+ \cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ x+ 1$

and I would attempt to prove that by "induction"

When n= 0, that says that y0= 1 which is true.

Now assume that, for some k, $\displaystyle yk= (1/k!)x^k+ (2/(k-1)!)x^{k-1}+\cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ x+ 1$ so that

$\displaystyle y(k+1)= \int x+ yk dx= \int (1/k!)x^k+ (2/(k-1)!)x^{k-1}+ \cdot\cdot\cdot+ (2/3!)x^3+ (2/2!)x^2+ 2x+ 1 dx$$\displaystyle = (1/(k+1)!)x^{k+1}+ (2/k!)x^k+ \cdot\cdot\cdot+ (2/4!)x^4+ (2/3!)x^3+ x^2+ x+ C$ and $\displaystyle y(k+1)(0)= C= 1$ gives

$\displaystyle y(k+1)= (1/(k+1)!)x^{k+1}+ (2/k!)x^k+ \cdot\cdot\cdot+ (2/4!)x^4+ (2/3!)x^3+ (2/2!) x^2+ x+ 1$

just as we wish.