Prove $99^{100}>100^{99}$ using binomial theorem

Question

Prove $0<(1+\frac{1}{n})<3$ and hence prove $99^{100}>100^{99}$.

I did the first part and showed $0<\frac{1}{n^{n-1}}\le3$ and hence $0<(1+\frac{1}{n})<3$. But for the second bit, I don't know how to incorporate the first bit to help me prove th inequality.

Presumably you mean $0 < \big(1+\frac1n\big)^n < 3$? – Mathmo123 Sep 09 '14 at 00:06 — Mathmo123, Sep 09 '14 at 00:06

score 5 · Accepted Answer · answered Sep 09 '14 at 00:04

5

Hint: Let $n = 99$. Then you have $$100^{99}=(1 +n)^n=n^n\left (1+\frac1n\right)^n$$

answered Sep 09 '14 at 00:04

Mathmo123

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Prove $99^{100}>100^{99}$ using binomial theorem

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