How can I prove that:
- $\forall x \in \mathbb{N}\setminus {0} \quad \dfrac{-1}{n+1}\le \int_0^1\dfrac{(-x)^n}{1+x} dx \le \dfrac{1}{1+n}$
- $\lim_{n\to+\infty}\Sigma_{i=1}^{n}\dfrac{(-1)^{i-1}}{i}$.
I've already proved that: $\int_0^1\dfrac{(-x)^n}{1+x}dx=\ln(2)-(1-\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{(-1)^{n-1}}{n})$
Hint
$$0\leq\frac{1}{1+x}\leq 1,\qquad 0\leq x\leq 1$$
$$-\int_0^1 dx=-1\leq\int_0^1 (-x)^n dx\leq \int_0^1 (x)^n dx=\frac{1}{n+1}$$
Moreover $$\sum_{i=1}^{n}\dfrac{(-1)^{i-1}}{i}=\sum_{i=1}^n (-1)^{i-1}\int_0^1 t^{i-1}dt=\int_0^1\sum_{i=1}^n(-t)^{i-1}=\int_0^1\frac{dt}{1+t}-\int_0^1\frac{(-t)^n}{1+t}$$ and use the first result to prove that the last integral converges to $0$ so we find $$\sum_{i=1}^\infty \dfrac{(-1)^{i-1}}{i}=\int_0^1\frac{dt}{1+t}=\log 2$$