Prove that I(n) is a strictly decreasing series: I(n)= $\int_0^1 \frac{(x^n)}{(x+1)} dx$ Prove that $I(n)$ is a strictly decreasing series:
$$I(n)=  \int_0^1 \frac{(x^n)}{(x+1)} dx.$$
If it were just decreasing I would have used the fact that when $x$ is between $0$ and $1$ ($0$ and $1$ are included), $x^n \geq x^{n+1}$. Then divide by $x+1$ and integrate. But I found the version of the same problem where it states that it has to be strictly decreasing. Any ideas?

The Mean Value Theorem for Integrals states

If $f$ is continuous and $g$ is positive and integrable, then there exists $c \in (a,b)$ such that
  $$ \int_a^b f(x)g(x) \, dx = f(c) \int_a^b g(x) \, dx. $$

Applying this with $f(x)=x$, $g(x) = \frac{x^n}{x+1}$ gives
$$ \int_0^1 \frac{x^{n+1}}{x+1} \, dx = c\int_0^1 \frac{x^n}{x+1} \, dx, $$
where $0<c<1$, which is what we wanted.
