Which term is bigger? $\sqrt[102]{101}$ or $\sqrt[100]{100}$

Question

Which term is bigger? $\sqrt[102]{101}$ or $\sqrt[100]{100}$

I tried AM-GM but didn't succeed.

Answer 1

If we raise both sides of $\sqrt[102]{101}$ and $\sqrt[100]{100}$ to the $100$ power we are then comparing $101^{100/102}$ and $100$. Or in other words $101^{50/51}$ and $100$. Raise both to the $51$ power and we're comparing $101^{50}$ and $100^{51}$.

Thus it's equivalent to compare $101^{50}$ and $100^{51}$.

Thus it's equivalent to compare $\left(\frac{101}{100}\right)^{50}$ and $100$.

Now $\left(\frac{101}{100}\right)^{50}=(1+\frac1{100})^{50}$

Using the binomial theorem this is

$$\sum_{n=0}^{50}{50\choose n}\left(\frac{1}{100}\right)^{50-n}$$

Which is

$$\sum_{n=0}^{50}\frac{50!}{n!(50-n)!}\left(\frac{1}{100}\right)^{50-n}$$

$$=\sum_{n=0}^{50}\frac{50\cdot49\cdots(n+1)}{(50-n)!}\left(\frac{1}{100}\right)^{50-n}$$

It is clear that $\frac{50\cdot49\cdots(n+1)}{100^{50-n}}<1$ since the top is $50-n$ numbers less than $100$ and the bottom is $50-n$ copies of $100$.

Thus $$=\sum_{n=0}^{50}\frac{50\cdot49\cdots(n+1)}{(50-n)!}\left(\frac{1}{100}\right)^{50-n}$$ $$<50 < 100$$.

Thus $101^{50}$ is less than $100^{51}$. Thus $\sqrt[102]{101}$ is less than $\sqrt[100]{100}$

Answer 2

$$\sqrt[102]{101}=(101)^{\frac{1}{102}}=\left(\left(101\right)^{\frac{25}{51}}\right)^{\frac{1}{50}}=\sqrt[50]{(101)^{\frac{25}{51}}}$$

$$\sqrt[100]{100}=\sqrt[50]{10}=(10)^{\frac{1}{50}}$$

So:

$$\sqrt[50]{(101)^{\frac{25}{51}}}<\sqrt[50]{10}$$ $$\sqrt[102]{101}<\sqrt[100]{100}$$

Answer 3

let $100=n$ then we have to prove that $$(n+1)^{1/(n+2)}<n^{1/n}$$ this is equivalent to $$\left(1+\frac{1}{n}\right)^n<n^2$$ this is true for $n=100$ since the left-hand side has the limit $e$

Answer 4

Let $R$ be the unknown relation. $R \in \{<,=,>\}$. $$101^{1/102}\, R \,100^{1/100}$$ $$101^{100}\, R \,100^{102} $$ We did this by raising both sides to the $102 \times 100$.

$$(1+100)^{100} = 100^{100}(\frac{1}{100}+1)^{100} < 100^{100}e$$

When compared to $100^{102}$, the answer is clear.

Answer 5

$\sqrt[100]{100}>\sqrt[102]{101}\iff 100^{102}>101^{100}\iff 100^{51}>101^{50}$

$\iff 100>\left(\frac{101}{100}\right)^{50}$. This inequality gives for $x,r>0$ that $(1+x)^r\le e^{rx}$, so

$\left(1+0.01\right)^{50}\le e^{50\cdot 0.01}=\sqrt{e}\ll 100$.

Proof of the inequality is simple: $\left((1+x)^{\frac{1}{x}}\right)^{rx}\le e^{rx}$.

Answer 6

$101^{50} = \sum_{k=0}^{50} \binom{50}{k} 100^{50-k}$.

${101^{50} \over 100^{51} } = {1 \over 100} \sum_{k=0}^{50} \binom{50}{k} {1 \over 100^k} = {1 \over 100} (1+ {1 \over 100})^{50} = {1 \over 100} (1+ {{1 \over 2} \over 50})^{50} \le {1 \over 100} \sqrt{e} <1$.

Answer 7

Consider function $f(x)=x^{1/x}$, $f'(x)=x^{1/x-2}(1-\ln(x))$, so function $f$ is decreasing on the interval $(\mathrm{e},\infty)$, thus $100^{1/100}>101^{1/101}>101^{1/102}$.