How do we know that $\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}< n $?

Question

I'm trying to convince myself that $$b^n - a^n < (b-a)nb^{n-1}$$ when $0 < a < b$ and $n>1$.

I will be persuaded once I can show that $$\sum\limits_{k=1}^nb^{n-k}a^{k-1}=b^{n-1}\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}< nb^{n-1},$$ or, more specifically, that $$\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}< n.$$

It sort of makes sense on a conceptual level, since I'm adding smaller and smaller terms, but for all I know, the summation on the left represents a bunch of small terms combining to a term bigger than $n$. Can someone help me definitively prove this?

Answer 1

First, $\left(\frac{a}{b}\right)^{k-1}=1$ for $k =1$.

Since $0<\frac{a}{b}<1$, $0<\left(\frac{a}{b}\right)^{k-1}<1$ for $k>1$.

Adding up $n$ such terms yields a sum smaller than $n$.

Answer 2

Obvious for $n = 2$. Assume it is true for $n$, let's prove it for $n+1$. $$\sum_{k=1}^{n+1}(\frac{a}{b})^{k-1}=\sum_{k=1}^{n}(\frac{a}{b})^{k-1} + (\frac{a}{b})^{n} < n + (\frac{a}{b})^{n}$$ But $(\frac{a}{b})^{n} < 1$ since $\frac{a}{b} < 1$. So $$\sum_{k=1}^{n+1}(\frac{a}{b})^{k-1} < n + 1$$

Answer 3

Alt. hint:   if calculus is allowed, just use MVT for $\,f(x)=x^n\,$, so that $\;\frac{b^n-a^n}{b-a} = n c^{n-1} \lt nb^{n-1}$ where $\,c \in (a,b)\,$, and the latter inequality holds because $\,f'(x)=nx^{n-1}\,$ is increasing on $\Bbb R^+\,$.