How to prove that $a(1-a^N)/(1-a)<N$?

Question

I derive the expectation for a problem and get $$N-\frac{a(1-a^N)}{1-a}$$ where $0<a<1$, and $N>0$.

The physical meaning of this value determines it should be positive. But I intuitively think $\frac{a(1-a^N)}{1-a}$ might exceed $N$ when $a$ is close to $1$.

However, I have enumerated some numerical values of $a$, i.e., $0.99999$, $0.999999$ and find $$\frac{a(1-a^N)}{1-a}<N$$ (where $0<a<1$, and $N>0$) holds true, so I conjecture it is always true, but am not sure how to prove it.

Answer 1

When $N$ is a positive integer, and $0<a<1$, then $$\frac{a(1-a^N)}{1-a}=a+a^2+a^3+\cdots+a^N<1+1+1+\cdots+1=N.$$

Answer 2

Well, $1-a^N=(1-a)(1+a+a^2+a^3+...+a^{N-1})$, and each of the terms in the second bracket is less than or equal to 1, so $1-a^N\leq(1-a)N$

Therefore, $\frac{a(1-a^N)}{1-a}\leq\frac{a(1-a)N}{1-a}\leq Na<N$ as long as $0<a<1$

Does that help?