(1) Base case: Show that holds true. We have: .

Induction step: Show that holds true leads to the fact that holds true.  So the statement follows.

(2) This problem is interestingly tricky.

First we find the critical points of : If then (due to the fact that ), so as .

If then .

If then the found critical point . We have and , so if then is the global maximum. It is easy to see that , (due to the fact that ). So in this case, .

To summarize: (3) First we prove that (the sum notation here is a little bit tricky for , reference):

Base case: .

Induction step: So the statement follows.

Second we prove that . So the statement follows.

Here I don’t understand why the problem only requires but not for any ?

(4) We have: It is shown from (3) that , so: I don’t know why I didn’t use the fact given by the examination though… (Something wrong?)

Thanks to my labmate, maybe I was wrong. It is not so trivial to prove that (this comes from the fact that ), . Therefore: So: (5) We proved in (3). So:  When approaches infinity, approaches , so does given small enough (you can factor into a sum and argue that the subsum approaches when approaches infinity, so no much difference in value between and ). So and will cancel each other out when approaches infinity and given small enough .

In the case becomes large enough that makes the difference between and noticeable, also starts to approaches and makes both and approaches . So and will cancel each other out given big enough .

Therefore: As hinted by (6), I’m pretty sure though.

(6) We proved in (3) and we proved in (5). So: So the statement follows.