Does (1+1/n)^2n converge or diverge? And how to approach arctan x?

[appplejack]

Homework Statement

I need to find whether the sequence converges or diverges
1.(1+ 1/n)^2n


2.arctan x

Homework Equations

The Attempt at a Solution

1. My answer is wrong but I don't know how I can get the right answer
My answer
(1 + 1/n) ^2n becomes 1 because 1/n tends to zero and 2n become infinitely big and
1 ^ infinity is 1. So the answer is 1 but the answer key says e^2

2. I don't know how to approach it

 

[McAfee]

1. The limit of (1+ 1/n)^n as n goes towards infinity is e. Since the question asks (1+ 1/n)^2n the answer is e^2.

Just take the limit of (1+ 1/n)^2n as n goes towards infinity.

2. the limit as arctan(n) goes to inifinity is infinity, thus arctan(n) diverges.

 

[RoshanBBQ]

$$\lim_{n\rightarrow \infty} \left (\left ( 1+\frac{1}{n} \right )^n \right)^2$$
$$\left ( \lim_{n\rightarrow \infty} \left ( 1+\frac{1}{n} \right )^n \right)^2$$

So let's just ignore the squaring for now. Whatever we find, we will square it.

$$I =\lim_{n\rightarrow \infty} \left ( 1+\frac{1}{n} \right )^n$$

$$ln(I) =\lim_{n\rightarrow \infty} ln\left (\left ( 1+\frac{1}{n} \right )^n \right )$$
$$ln(I) =\lim_{n\rightarrow \infty} nln\left (1+\frac{1}{n} \right )$$

This is an indeterminate form of zero times infinity. Let
$$N = \frac{1}{n}$$
then
$$ln(I) =\lim_{n\rightarrow 0} \frac{ln\left (1+N \right )}{N}$$

Can you take it from here?

 

[Bohrok]

For number 2, have you looked at the graph of arctan x?