# Logarithmic equation proof

How do I prove that this statement is true?

$$\frac{1}{\log(\log(n))} \ge \frac{1}{\log(\log(n+1))}$$

I only have this.

$$\log(\log(n+1)) \ge \log(\log(n))$$

Sorry for such a stupid question, but I've forgotten almost everything about logarithmic equations.

• In general, if $0 < a < b$ then $0 < {1 \over b} < {1 \over a}$. Nov 20, 2016 at 19:50
• @copper.hat That's pretty much what he had after he said "I only have this." Nov 20, 2016 at 19:53
• @AlgorithmsX: That is all that is needed here. Nov 20, 2016 at 19:54
• Are you asking how to show that $\log( \log (n+1)) \ge \log( \log n)$ is true, or how to show the top equation given the second equation (because the latter is what you have asked)? Nov 20, 2016 at 19:55
• Note: the statement is not true for certain small $n$, e.g. for $n=2$, $\frac{1}{\log(\log(2))}<0$ while $\frac{1}{\log(\log(3))}>0$. Nov 20, 2016 at 20:00

$$n\le n+1\\ \log { n } \le \log { \left( n+1 \right) } \\ \log { \left( \log { n } \right) \le \log { \left( \log { \left( n+1 \right) } \right) } } \\ \frac { 1 }{ \log { \left( \log { n } \right) } } \ge \frac { 1 }{ \log { \left( \log { \left( n+1 \right) } \right) } }$$

The inequation holds when both sides are the same sign

• You should note in this answer that this reasoning only works if both of the two $\log(\log)$ terms are of the same sign. Nov 21, 2016 at 5:41

Hint:

Logarithmic functions are monotonically increasing on their domain. That is, if $a>b$, then $\log a > \log b$.

Additionally, for positive $a$ and $b$, if $a>b$, then $1/a<1/b$.

### TL;DR

If and only if you are given that either

$$0 \lt \log{(\log{(n)})} \le \log{(\log{(n+1)})}$$

Or

$$\log{(\log{(n)})} \le \log{(\log{(n+1)})} \lt 0$$

$$\frac {1}{\log{(\log{(n)})}} \ge \frac {1}{\log{(\log{(n+1)})}}$$

Since the inverse function is strictly decreasing in the positive domain and in the negative one.

### (Not too) long version

Actually, given what you have:

$$\log{\log{(n)}} \le \log{(\log{(n+1)})}$$

It is not always true that

$$\frac {1}{\log{(\log{(n)})}} \ge \frac {1}{\log{(\log{(n+1)})}}$$

Especially that you did not specify the base you are using for your logarithm function. For example, if you are using the natural logarithm (i.e. "base-$e$") then take $n=2$:

$$\log{(n)} = \ln{(2)} \approx 0.693$$ $$\log{(n+1)} = \ln{(3)} \approx 1.099$$

$$\log{(\log{(n)})} = \ln{(\ln{(2)})} \approx \ln{(0.693)} \approx -0.367$$ $$\log{(\log{(n+1)})} = \ln{(\ln{(3)})} \approx \ln{(1.099)} \approx 0.094$$

You can see where this is going! Sure, what you were given still holds, namely that $\log{(\log{(n)})} \le \log{(\log{(n+1)})}$, but applying the inverse function at this point would actually preserve the order, since what you really have is

$$\log{(\log{(n)})} \lt 0 \lt \log{(\log{(n+1)})}$$

$$\frac {1}{\log{(\log{(n)})}} \lt 0 \lt \frac {1}{\log{(\log{(n+1)})}}$$

Because the inverse function is only strictly decreasing on each side of the y-axis separately, whereas every point on the right side is strictly greater than every point on the left side (inverse function from -4.2 to 4.2).

The point from all of this is highlighted in the TL;DR section: you need to know beforehand that $\log{(\log{(n)})}$ and $\log{(\log{(n+1)})}$ are either both strictly positive or strictly negative for all of this to work.

Now if the base of your logarithm function is an integer, then you won't have this problem since no two consecutive integers (assuming you are indeed using $n$ to indicate an integer) applied twice to an integer-base logarithm function, give two results with opposite signs. However, one of the results can be $0$ in which case the inverse is not even defined...

$$e^{e^{\log(\log(n+1))}}=e^{\log(n+1)}=n+1$$ You should be able to take it from there.

log(x) is a monotonically increasing function

1/x is a monotonically decreasing function

decreasing(increasing(increasing)) = decreasing