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?

Hint:
consider $f(t) = \dfrac{t}{1-x^t}$ for $|x| < 1$. It would be almost the same as your idea.
