Prove that: $\lfloor n^{1/2}\rfloor+\cdots+\lfloor n^{1/n}\rfloor=\lfloor \log_2n\rfloor +\cdots+\lfloor \log_nn \rfloor$, for $n > 1$

Question

Prove that: $\lfloor n^{1/2}\rfloor+\cdots+\lfloor n^{1/n}\rfloor=\lfloor \log_2n\rfloor +\cdots+\lfloor \log_nn \rfloor$, for $n > 1,\, n\in \mathbb{N}$

For example. For $n=2$, we have $\lfloor 2^{1/2} \rfloor = \lfloor 1.414 \rfloor = 1$ whereas $\lfloor \log_2(2) \rfloor = 1$ while for $n=3$, we have $$\lfloor 3^{1/2} \rfloor + \lfloor 3^{1/3} \rfloor = \lfloor 1.732 \rfloor + \lfloor 1.442 \rfloor = 2= \lfloor 1.585 \rfloor + \lfloor 1 \rfloor=\lfloor \log_2(3) \rfloor + \lfloor \log_3(3) \rfloor .$$

I was thinking of using induction. So since $n=2$ is true, now assume for all $n$, this identity is true, we would like to prove that $n+1$ is true. Then

$$\lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor + ... + \lfloor n^{1/n} \rfloor + \lfloor (n+1)^{1/(n+1)} \rfloor,$$ where $(n+1)^{1/(n+1)} > 1$ for all $n>1$ but it's strictly decreasing above 1 so $\lfloor (n+1)^{1/(n+1)} \rfloor = 1$

$\implies \lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor +\cdots+ \lfloor n^{1/n} \rfloor + \lfloor (n+1)^{1/(n+1)} \rfloor = \lfloor n^{1/2} \rfloor + \lfloor n^{1/3} \rfloor +\cdots+ \lfloor n^{1/n} \rfloor + 1 $

$= \lfloor \log_2(n) \rfloor + \lfloor \log_3(n) \rfloor + \cdots+ \lfloor \log_n(n) \rfloor + \lfloor \log_{n+1}(n+1) \rfloor$

since, $\log_{n+1}(n+1) = 1$ for all $n$.

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My question is: How do we know that $(n+1)^{1/(n+1)}$ will never go below $1$? i.e., How can we prove that this function $f(x) = (x+1)^{1/(x+1)}$ is always bounded below by $1$ for $x>1$? (First, When $x=0$, $f(0)=1$, then looking at it's derivative, one can see that it's strictly increasing for $x$ between $(0,1)$ and decreasing for all $x>1$).

Answer 1

This is a classic exercise and one with a very elegant solution.

The idea of the proof is to count the number $N$ of the points (see figure below) with integer coordinates, which lie in the region $$ U=\big\{(x,y): 0<x\le n \,\,\,\text{and}\,\,\, 1<y\le n^{1/x}\big\}, $$ and in particular, the red points, in two ways: horizontally and vertically.

Horizontal counting: $$ N=\lfloor n^{1/2}\rfloor+\lfloor n^{1/3}\rfloor+\cdots+\lfloor n^{1/n}\rfloor, $$ since on the horizontal line $\,y=k\,$ lie exactly $\,\lfloor n^{1/k}\rfloor\,$ red points.

Vertical counting: $$ N=\lfloor \log_2 n\rfloor+\lfloor\log_3 n\rfloor+\cdots+\lfloor \log_n n\rfloor, $$ since on the vertical line $\,x=k\,$ lie exactly $\,\lfloor \log_k n\rfloor\,$ red points.

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Note that the curve in the figure above is of the function $y=n^{1/x}$.

This problem was first asked in a Soviet Mathematics Olympiad in 1982 (Всесоюзный Математический Олимпиад.)

Answer 2

Fix $b>1$. Then the derivative of $b^x$ is $\ln(b) b^x$; $\ln(b)$ is positive and $b^x$ is as well for all $x$, showing that that $b^x$ is a strictly increasing function. Next, $b^0=1$, showing that $b^x>1$ for all $x>0$.

Next, since $n+1>1$ and $1/(n+1)>0$, we have that $(n+1)^{1/(n+1)}>1$.