The latter formula of the limit can be transformed into [(n+2)/(n+1)] [n+1] [n/(n+1)] n [(n+2)/(n+65438).

These three items are limited respectively, and the first two items are mainly used.

Lim(n tends to infinity) (1+1/n) n = e.

The basic formula for finding the limit. Note that the condition is that n tends to infinity, that is, it contains both positive infinity and negative infinity.

So we can get the first two cracks.

[1+1(n+1)] (n+1), limit e.

[[1+1/(-(n+1))] (-n)] (-1), the limit is1/e.

Finally, the limit of the three terms can be calculated by combining the limit algorithm, and then multiplied.

The second problem can also be summarized by the split term method, which can be reduced to

an= 1/[n！ (n+2)]=(n+ 1)/(n+2)！

I suddenly lost my train of thought here, but I was thinking that this problem could only be solved, so I tried it, and the formula above is exactly equal to

1//(n+ 1)！ - 1/(n+2)！

So the answer is1/2-1102!