series of series of integrals, where it can converge?

$f_1(x)=\frac{M}{C}$, where M and C are constants $$f_2(x)=\frac{\int_0^xf_1(y)dy}{C} + \frac{\int_0^x\int_0^yf_1(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_1(t)dtdzdy}{C^3} + \cdots$$ $$f_3(x)=\frac{\int_0^xf_2(y)dy}{C} + \frac{\int_0^x\int_0^yf_2(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_2(t)dtdzdy}{C^3} + \cdots$$

$$f_{i+1}(x)=\frac{\int_0^xf_i(y)dy}{C} + \frac{\int_0^x\int_0^yf_i(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_i(t)dtdzdy}{C^3} + \cdots$$

$$f_{\infty}(x) = ?$$

I want to examine the convergence of the function of $f_{\infty}(x)$.
Each function $f_i(x)$ can be represented with $e^{\frac{x}{C}}$ function as a shorter version by using the maclaurin Series.

$f_2(x)$ becomes $\frac{M}{C} \left[ e^{\frac{x}{c}}-1 \right]$ when the infinite series is arranged by using the Maclaurin Series.
$f_3(x)$ also becomes $\frac{M}{C} \left[ e^{\frac{x}{c}} \left( e - \frac{x}{c} \right) - e \right]$.
So far I calculated $f_4(x)$, which has too many terms to be written down here.
By writing program codes, I calculated and found that $f_i(x)$ function is getting closer to the function $\frac{2M}{C^2}x$ with increasing i.
I want to mathematically prove this convergence. $$f_{\infty}(x)=\frac{2M}{C^2}x$$ Any tip will be appreciated.

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Your integrals are not well defined ; instead of putting $x$'s everywhere you should put different variables for the upper bounds of the integrals and the $dx$'s. Please modify your text to make it more understandable. –  Patrick Da Silva Mar 13 '12 at 8:04
@DidierPiau you are right... thank you for the comment.. I've modified it. –  Nate Mar 13 '12 at 8:32
@PatrickDaSilva Thank you for the comment... I've modified it... –  Nate Mar 13 '12 at 8:36
Yes, that is better. =) –  Patrick Da Silva Mar 13 '12 at 16:32

For every $i\geqslant0$ and $x\geqslant0$, $$f_i(x)=\frac{M}C\mathrm e^{x/C}\sum_{k=i}^{+\infty}\frac{(-1)^{k-i}}{C^k}\frac{x^k}{k!}.$$ Hence $f_i(x)\to0$ for every $x\geqslant0$.

To prove the convergence, fix $x_i\gt0$, $i\geqslant3x_i/C$ and some $K_i$ large enough such that $f_i(x)\leqslant K_ix^i$ for every $x\leqslant x_i$. Such a number $K_i$ exists for every $x_i$ because $f_i(x)\sim Mx^i/(C^{i+1}i!)$ when $x\to0$ and $f_i$ is continuous.

Then, for every $j\geqslant i$, $Cf_{j+1}'=f_j+f_{j+1}$ and $f_{j+1}(0)=0$ yield $f_{j+1}(x)\leqslant K_{j+1}x^{j+1}$ for every $x\leqslant x_i$, with $$K_{j+1}=\frac{K_j}{C\cdot(j+1)-x_i}\leqslant\frac{K_j}{2x_i}.$$ Hence, for every $j\geqslant i$, $K_j\leqslant K_i/(2x_i)^{j-i}$. This implies that $f_j(x)\leqslant K_jx^j\leqslant K_ix^i/2^{j-i}$ for every $x\leqslant x_i$, hence $f_j\to0$ on $(0,x_i)$. Since $x_i$ is as large as desired, the proof is complete.

Note that the limit equation, whatever that means, is $Cf_{\infty}'=2f_\infty$, whose solutions are $f_\infty(x)=K_\infty\mathrm e^{2x/C}$ with $K_\infty=f_\infty(0)$. The convergence above is simply the translation of the fact that $K_\infty=0$.

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Thank you very much for the answer. I don't know how to thank you enough... Actually I've made a mistake in simplifying the equations before asking this question. Later I found that and I tried to fix it and to edit this page but at that moment you've already answered this question. A few minute ago, I posted the fixed question which was what I originally intended to ask. However, I think that this answer will be also helpful for me to figure out how to solve the fixed problem. Thank you again. –  Nate Mar 13 '12 at 12:06