Prove that $\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}$ given $n$ is a positive integer and $0 < x \lt 1$. Problem Statement: If $n$ is a positive integer and $0 < x \lt 1$, show that
$$ \frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}.$$
My Solution:
$$
\frac{ 1- x^{n+1} }{n+1} \lt \frac{1-x^n}{n} ~~~~\text{is true} \\
\text{if}~~~~ \frac{n}{1-x^n} \lt \frac{n+1}{1- x^{n+1} }$$
If we see the LHS we find that it is of the form of sum of a geometric series with the first term $n$ and the common ration $x^n$ (which is less than 1) similarly the RHS represents the sum of a geometric series with the first term as $n+1$ and common ratio $x^{n+1}$ (which is less than $x^n$, that is less than 1)
Now, my point is that the series represented by the LHS have the first term lesser than the first term of the series represented by the RHS, and the series represented by LHS decreases fastly in comparison to the series represented by RHS (because the common ratio $x^{n+1} \lt x^n$), hence the sum of series represented by the RHS is greater than the sum of the series represented by the LHS.
Is my solution and reasoning correct?
 A: Another option:
$$
 \frac{1-x^n}{n} = \int_x^1 t^{n-1} \, dt
$$
and the integrand $t^{n-1}$ decreases strictly on $(x, 1)$ if $n$ increases.

Yet another option: The inequality
$$
\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}
$$
is equivalent to
$$
x^n < \frac{1}{n+1}x^0 + \frac{n}{n+1}x^{n+1}
$$
and that is true because the function $f(t) = x^t$ is strictly convex for fixed $x \in (0, 1)$. Graphically: The slope of the secant
$$
 \frac{x^n-1}{n} = \frac{f(n)-f(0)}{n-0}
$$
increases with increasing $n$.
A: By your reasoning we need to prove that:
$$n(1+x^n+x^{2n}+...)<(n+1)(1+x^{n+1}+x^{2(n+1)}+...)$$ and it's not so clear, why it's true.
Another way:
We need to prove that:
$$nx^{n+1}-(n+1)x^n+1>0,$$ which is true by AM-GM:
$$nx^{n+1}+1\geq(n+1)\sqrt[n+1]{\left(x^{n+1}\right)^n\cdot1}=(n+1)x^n.$$
The equality occurs maybe for $x^{n+1}=1,$ id est, does not occur.
A: Hint:
consider $f(t) = \dfrac{t}{1-x^t}$ for $|x| < 1$. It would be almost the same as your idea.
A: We have that
$$\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n} \iff n-nx^{n+1} \lt n-nx^n+1-x^n$$
$$\iff nx^n(1-x) \lt 1-x^n \iff nx^n<\overbrace{1+x+x^2+\ldots+x^{n-1}}^{\color{red}{\text{n terms}\,> \,x^n}}$$
A: 
Is my solution and reasoning correct?

No, although the first term of the LHS is smaller than the second, its not true that the terms of the implied LHS geometric series have a smaller common ratio than those of the RHS. If the common ratio is smaller, then I would say that it "decreases faster", but the smaller common ratio, $x^{n+1}$ is on the right. What your argument proves is the slightly less interesting
$$ \frac n{1-x^{n+1}} < \frac{n+1}{1-x^n}$$
and your paragraph of words is a manifestation of the rule "$a\le b, \ c< d \implies ac < bd$". (Indeed,  $$\frac1{1-x^{n+1}} < \frac1{1-x^n} \iff 1-x^n < 1-x^{n+1} \iff x^{n+1} < x^n \iff x< 1$$  by the assumption $x\in(0,1)$.) The result you are tasked to prove does not follow from this rule.
