Logarithmic equation proof

Question

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.

Answer 1

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

The inequation holds when both sides are the same sign

Answer 2

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

Answer 3

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 $1/a<1/b$.

Answer 4

TL;DR

If and only if you are given that either

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

Or

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

Then you can simply jump to the conclusion:

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

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 $n=2$:

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

Which leads to

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

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

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

Which leads to

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

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...

Answer 5

log(x) is a monotonically increasing function

1/x is a monotonically decreasing function

decreasing(increasing(increasing)) = decreasing