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I'm stuck with induction problem. Let $0\leq a<b$, prove the next inequality for all $n\in\mathbb{N}$.

$$\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$$I need this inequality because with this, we can deduce that the sequence $\left(1+\frac{1}{n} \right)^{n}$ converges.

The base, for $n=1$ it's trivial because $\displaystyle\frac{b^2-a^2}{b-a}=b+a<2b$ because by hypothesis $a<b$. Next suposse that for $n=k$ the inequality holds. If we take the inequality then $$\frac{b^{k+1}-a^{k+1}}{b-a}<(k+1)b^k$$ and multiplying for $b$ in both sides we obtain $$\frac{b^{k+2}-ba^{k+1}}{b-a}<(k+1)b^{k+1}<(k+2)b^{k+1}$$but from here I don't know how to obtain the result. Any hint? Thanks for your help.

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Factor out $\frac{b^{n+1}}{b}=b^n$ and get

$$b^n \left(\frac{1-\left(\frac{a}{b}\right)^{n+1}}{1-\frac{a}{b}}\right)=b^n\sum_{k=0}^n\left(\frac{a}{b}\right)^k$$

by the geometric sum formula. Then since $0\leq a<b \implies 0 \leq \frac{a}{b} < 1$ we have for $n>0$

$$b^n\sum_{k=0}^n\left(\frac{a}{b}\right)^k < b^n\sum_{k=0}^n 1 = (n+1)b^n$$

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