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Let $a_n = n^{1/n}$ and $b_n = \dfrac{a_{n+1}}{a_n}$. I am trying to prove that $\{b_n\}_{n=5}^{\infty}$ is increasing. Here is what I know and what I've tried:

  • The first few elements of $b_n$ aren't increasing (because $a_n$ is decreasing only after $n \geq 3$, for instance.) and so it wouldn't be correct to prove that $b_n$ is increasing for all $n$. Then letting $S(n)$ be the statement $$b_{n} = \frac{(n+1)^{1/(n+1)}}{n^{1/n}} \leq \frac{(n+2)^{1/(n+2)}}{(n+1)^{1/(n+1)}} = b_{n+1}$$ we can assume $S(n-1)$ and prove $S(n)$ (and $S(5)$). I tried to do this by breaking apart the exponents and manipulating the inequalities but I couldn't get anywhere.

  • The other thing I tried was to prove that $$a_n^{2} \leq a_{n-1}a_{n+1}$$ using a convexity argument, particularly if $f(n) = n^{1/n}$ that $$a_n = f(n) = f(\frac{(n-1) + (n+1)}{2}) \leq \frac{f(n-1) + f(n+1)}{2} = \frac{a_{n-1} + a_{n+1}}{2}$$ I figured the AM-GM inequality might be useful because of the RHS but it implies that $$a_{n-1}a_{n+1} \leq (\frac{a_{n-1} + a_{n+1}}{2})^2$$ which doesn't seem to help bound $a_n^2$ above.

This question is identical to mine but I'm looking for an approach that doesn't make use of logarithms or differentiation, since the book I'm working through hasn't covered those yet.

If I could have a small hint to point me in the right direction I would appreciate it!

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It is no hard to prove $(1+\dfrac1n)^n$ is increasing and $(1+\dfrac1n)^n<e$, then $$\left(\dfrac{b_{n+1}}{b_n}\right)^{n(n+1)(n+2)}=\left(\dfrac{a_{n+2}a_{n}}{a_{n+1}^2}\right)^{n(n+1)(n+2)}=\dfrac{(n+2)^{n(n+1)}n^{(n+1)(n+2)}}{(n+1)^{2n(n+2)}}=$$ $$\dfrac{\left(\dfrac{n+2}{n+1}\right)^{n(n+1)}}{\left(\dfrac{n+1}{n}\right)^{n(n+3)}}n^2 = \left(\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{\left(1+\dfrac{1}{n}\right)^{n}}\right)^n\dfrac{n}{\left((1+\dfrac1n)^n\right)^\frac32}>\dfrac{n}{e^\frac32}>1$$ valid if $n>e^\frac32>4$.

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  • $\begingroup$ Interesting that you came up with the same bound ($n > e^{3/2}$) that I did. I think that your answer is much better, also. $\endgroup$ – marty cohen Sep 2 '18 at 16:31
  • $\begingroup$ @martycohen thank you. $\endgroup$ – Nosrati Sep 2 '18 at 16:34
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$a_n = n^{1/n} $ and $b_n = \frac{a_{n+1}}{a_n} $.

Let $c_n =\ln(a_n) =\dfrac{ln(n)}{n} $ and

$\begin{array}\\ d_n &=\ln(b_n)\\ &=\ln(a_{n+1})-\ln(a_n)\\ &=\dfrac{ln(n+1)}{n+1}-\dfrac{ln(n)}{n}\\ &=\dfrac{nln(n+1)-(n+1)\ln(n)}{n(n+1)}\\ &=\dfrac{n(\ln(n)+ln(1+1/n))-n\ln(n)-\ln(n)}{n(n+1)}\\ &=\dfrac{n\ln(n)+nln(1+1/n)-n\ln(n)-\ln(n)}{n(n+1)}\\ &=\dfrac{nln(1+1/n)-\ln(n)}{n(n+1)}\\ &=\dfrac{ln(1+1/n)}{n+1}-\dfrac{\ln(n)}{n(n+1)}\\ \text{so}\\ d_n' &= \dfrac{(2 n + 1) (\ln(n) - 1) - n^2 \ln(1/n + 1)}{n^2 (n + 1)^2}\\ \end{array} $

If we can show that

$e_n =(2 n + 1) (\ln(n) - 1) - n^2 \ln(1/n + 1) \gt 0$ for large enough $n$ (Wolfy says $n > 4.05$), we are done.

$n^2 \ln(1/n + 1) \lt n^2\dfrac1{n} =n$ so

$\begin{array}\\ e_n &=(2 n + 1) (\ln(n) - 1) - n^2 \ln(1/n + 1)\\ &\gt (2 n + 1) (\ln(n) - 1)-n\\ &= (2 n + 1) \ln(n) - (2n+1)-n\\ &= (2 n + 1) \ln(n) - (3n+1)\\ &= (2 n + 1) \left(\ln(n) - \dfrac{3n+1}{2n+1}\right)\\ &= (2 n + 1) \left(\ln(n) - \left(\dfrac{3n+1}{2n+1}-\dfrac32+\dfrac32\right)\right)\\ &= (2 n + 1) \left(\ln(n) - \left(\dfrac{(3n+1)2-(2n+1)3}{2(2n+1)}+\dfrac32\right)\right)\\ &= (2 n + 1) \left(\ln(n)-\dfrac32+ \dfrac{1}{2(2n+1)}\right)\\ &= (2 n + 1) \left(\ln(n)-\dfrac32\right)+\dfrac12\\ \end{array} $

Therefore, if $\ln(n) \gt \dfrac32$, or $n > e^{3/2} \approx 4.481$. this is true.

Therefore this is true for $n \ge 5$.

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  • $\begingroup$ From the question: ".. but I'm looking for an approach that doesn't make use of logarithms or differentiation, since the book I'm working through hasn't covered those yet." $\endgroup$ – Martin R Sep 2 '18 at 6:31

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