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I am given a sequence $(f_n)_n$ where $n\in N$. $f_n : \Re \rightarrow \Re: x \mapsto 1$

$f_1:\Re \rightarrow \Re$ is defined as follows

$$f_1 (x) = 1 + \int_0^x f_0 (t) dt$$

One sees that the sequence has to be:

$$f_{n+1} (x) = 1 + \int_0^x f_n (t) dt$$

Now I want to compute the value of these functions. One gets:

For $n = 0$

$$f_{1} (x) = 1 + \int_0^x f_0 (t) dt$$

(the provided one).

For $n = 1$

$$f_{2} (x) = 1 + \int_0^x f_1 (t) dt$$

For $n = 2$

$$f_{3} (x) = 1 + \int_0^x f_2 (t) dt$$

$$...$$

Basically what I am doing here is iterating and what I have finally got is:

Knowing that $f_n : \Re \rightarrow \Re: x \mapsto 1$

$$f_{1} (x) = 1 + x$$

$$f_{2} (x) = 1 + x + \frac{x^2}{2}$$

$$f_{2} (x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$

$$...$$

Thus the sequence $(f_n)_n$ does not seem to converge while it should...

What's wrong with my reasoning?

I am also curious on how could I prove the convergence of a sequence which includes an integral on it. I have been studying the convergence of a sequence theorem and understood it. I have applied it to straightforward examples but never to a sequence with an integral on it. How could I do so?

EDIT:

So we get that

$$f_{n+1} (x) = \sum_{n}^{N} x^{n}/n!$$

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  • $\begingroup$ Redo te computation for $f_2$, you should have $x^2/2$. $\endgroup$ – Mindlack Feb 11 at 18:14
  • $\begingroup$ @Mindlack why, isn't $f_0 = 1$? $\endgroup$ – JD_PM Feb 11 at 18:17
  • $\begingroup$ Sure it is. And $f_1(x)=1+x$ as well. But what is $f_2$? $\endgroup$ – Mindlack Feb 11 at 18:19
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    $\begingroup$ Look at fixed point operators. If $(Af)(x) = 1+\int_0^x f(t)dt$, then the iteration is $f_{n+1} = A f_n$ and you are (presumably) trying to show that $f_n$ converges to the fixed point $f$ that satisfies $f=Af$. $\endgroup$ – copper.hat Feb 11 at 18:29
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    $\begingroup$ No. $f_2(x)=1+\int_0^x{(1+t)\,dt}=1+x+x^2/2$. $\endgroup$ – Mindlack Feb 11 at 18:36

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