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Let $a_n = \sqrt[n]{n}$ and $b_n = a_{n+1}/a_n$. Prove that $\{b_n\}_{n=5}^\infty$ is increasing and find the limit.

I've tried to show that this sequence is increasing without luck. Specifically, I've looked at showing $b_{n+1} - b_n > 0$ or $\frac{b_{n+1}}{b_n} > 1$ but I can't show that the sequence is increasing with either inequality. Is this even the right approach?

mathjacks
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3 Answers3

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Since $\ln b_n=\frac{\ln(n+1)}{n+1}-\frac{\ln n}n$, define $f(x)=\frac{\ln x}{x}$. We need to prove that $f(x)$ is convex on $[5,+\infty]$. This follows from$$f'(x)=\frac{1-\ln x}{x^2}\Rightarrow f''(x)=\frac{2\ln x-3}{x^3}>0,$$for $\ln5=1.6...>\frac{3}{2}$.

Eclipse Sun
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$$ \log b_n = \frac{\log (n+1)}{n+1} - \frac{\log n}n \\ \log b_n - \log b_{n-1} = \frac{\log (n+1)}{n+1} - 2\frac{\log n}n + \frac{\log (n-1)}{n-1} $$

Let $f(x) = \frac{\log x}x$. Then: $$ f'(x) = \frac{1 - \log x}{x^2}\\ f''(x) = \frac{2\log x - 3}{x^3} \ge 0 $$ for $x>e^{3/2}\simeq 4.8$.

hence, using the mean value theorem: $$ f(x+1) - f(x) = 1\times f'(c_x) $$with $x< c_x < x+1$. Using the fact that $f'$ increases for $x\ge 5$, you get $$ \frac{\log n+1}{n+1} - 2\frac{\log n}n + \frac{\log n-1}{n-1} \ge 0 \\ \implies b_n \ge b_{n-1} $$

mookid
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Let : f(x) = $ \frac{ln(x)}{x} $

$f'(x) = \frac{1}{x^2} -\frac{ln(x)}{x^2} $

For x $\geq e$ f decreases, and f(x) ->0 when x-> $+\infty$

Now, $ a_n = e^{f(n)} $ => $(a_n) -> 1$ , when $ n-> +\infty $

So without studying the monotony of ($b_n$), you can say that ($b_n$) -> 1, when $ n->+\infty$

mvggz
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