Let's start with question 3: Because lim (a_i)/n=0, the limit is split.
Let's talk about the problem 1: the existence and equality of any subsequence limit can definitely prove the existence of sequence limit. But we can't take any subsequence when we prove the limit, so we have a classic example:
lim a_{2k}=lim a_{2k+ 1}=a？ lim a_n=a。
You may have encountered a classic proof problem:
lim a _ { 3k } = lim a _ { 3k+ 1 } = lim a _ { 3k+2 } = a？ lim a_n=a
That is to say, it is not necessary to list all the sub-columns (and certainly not all of them) to prove the limit of series. From the above two examples, I personally got a small guess:
In {a_n}, for any positive integer p, {[0], [1], ... module p [p- 1]} converges, so {a_n} converges.
P=2 is an even-even column, and p=3 is {3k, 3k+ 1, 3k+2}, etc. It should not be difficult to prove (it is yours ...).
Look at question 2: If you look at the answer to question 3, question 2 is easy to understand, because this idea is essentially a congruence problem of number theory. . There is a positive integer p in the topic, and it is obvious that the congruence class of module p is a perfect division of n into several sub-columns.
In fact, some tricks in the competition are not all thoughtful, even if they are, they are also a kind of accumulation and vision. See more to see more. Let's go