# Convergence of continuous function

Let $f$ be a continuous function on [$a$,$b$] mapped onto $\mathbf{R}$ and which is differentiable at $c$ $\in$ ]a,b[.

(i)Show that $\exists$ a unique function $\epsilon$:]$a$,$b$[ mapped onto $\mathbf{R}$ $\ni$ $f(c)=0$,and that $\forall$ $x \in ]a,b[$, distinct from c, we have $f(x)=f(c)+(x-c)f'(c)+(x-c)\epsilon(c)$

(ii)Show that the sequence, $(s_n)_{n\ge 1}$, given by

$s_n=\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}$=$∑_{k=0}^n\frac{1}{(n+k)}$ is decreasing and that it converges to a limit $s$.

(iii)Why can we say that $\frac{1}{2}\le s \le 1$

(iv) Let f ]-1,1[ mapped onto $\mathbf{R}$ be continuous, differentiable at $x=0$ and $\ni$ $f(0)=0$.Show that the sequence ($\sigma_n (f))_{n\ge 1}$ given by

$\sigma_n(f)=f(\frac{1}{n})+f(\frac{1}{n+1})+...+f(\frac{1}{2n})$

converges to $f'(0)s$.

For the 1st part, I have started using Mean Value Theorem(MVT). $f(c)=0$, does this imply we should apply Rolles theorem here. $f'(c)(b-a)=f(b)-f(a)$

$f(c)=0$, does this imply we should apply Rolles theorem here?

For part (ii), what kind of sum should I use here.

The rest is a good understanding of the 1st part.

For part (ii):

Note that $s_n = H_{2n} - H_n + \frac1{n},$ where

$$H_n = \sum_{k=1}^n \frac1{k},$$

and

$$\frac1{k+1} < \int_k^{k+1} \frac{dt}{t} =\log(k+1) - \log k < \frac1{k}.$$

Summing from $k = 1$ to $k = n-1$ we get

$$H_n -1 < \log n < H_n - \frac1{n}.$$

Hence,

$$\log n + \frac1{n} < H_n < \log n +1, \\ \log 2 + \log n + \frac1{2n} < H_{2n} < \log 2 + \log n +1, \\ \log 2 - 1 + \frac1{2n} < H_{2n} - H_n < \log 2 + 1 - \frac1{n}.$$

Thus, $s_n$ is bounded as

$$\log 2 -1 < \log 2 - 1 + \frac1{2n} + \frac1{n} < s_n = H_{2n} - H_n + \frac1{n} < \log 2 + 1.$$

Now show $s_n$ is decreasing and, therefore, convergent by the monotone convergence theorem.

Note that

$$s_{n+1} - s_n = \frac1{2n+2} + \frac1{2n+1} - \frac1{n} = \frac{-3n - 2}{n(2n+1)(2n+2)} < 0$$