Question

How to prove \(X_n = \dfrac{n^2}{n+1}\) diverge?

Answer 1

using limit ?

Answer 2

Using limit theorem, not definition.

Answer 3

http://archives.math.utk.edu/visual.calculus/1/limits.18/

Answer 4

Limit theorems talk about convergent sequence only. That is the problem, :(

Answer 5

Use the fact that \(\dfrac{n^2}{n+1}=n-1+\dfrac{1}{n}\).

Answer 6

Yes, I did, but lim n = infinitive get 0 credit from my prof. :)

Answer 7

He said: you can break X_n into many parts as you like, but each part must be a convergent to apply limit theorem there. Your n is a divergent sequence and we never construct the formula lim n = infinitive.

Answer 8

try assuming there exists a limit L and arrive at a contradiction

Answer 9

\[\large \left|\frac{n^2}{n+1} - L\right|\lt \epsilon\]

Answer 10

or are you not allowed to use contradiction ?

Answer 11

lol he gave you negative marks :o

Answer 12

Yes. :)

Answer 13

can't you use L'Hopital's rule

Answer 14

:( No, I can't. Just limit theorem. even definition of limit is not allowed. :(

Answer 15

this page is next to the previous one. Just take a look for fun. :)

Answer 16

:S

Answer 17

\[\large \lim\limits_{n\to \infty} \dfrac{n^2}{n+1} =\lim\limits_{n\to \infty} \dfrac{n^2}{n} = \lim\limits_{n\to\infty} n =\infty \]

Answer 18

because as n->infty, n+1 ~ n

Answer 19

Ok, hopefully with this logic, I can get back 2 credits. hahaha..
No worry, I am already get A^- for this homework. :)

Answer 20

idk if there is a more pleasing proof @nerdguy2535

Answer 21

he is offline :)

Answer 22

@freckles @zepdrix @AccessDenied

Answer 23

You should able to prove just lim n-> inf n=inf
since
\[\frac{n^2}{n+1}=n+\frac{-1}{n+1}\]
and -1/(n+1) goes to 0

Answer 24

I think the idea that Sith had above can work, you just have to work it properly. We have that: $$\frac{n^2}{n+1}=n-1+\frac{1}{n}\iff n=\frac{n^2}{n+1}-\frac{1}{n}+1.$$We proceed by contradiction. Assume that: $$\lim_{n\rightarrow \infty}\frac{n^2}{n+1}=L.$$ Since: $$\lim_{n\rightarrow \infty }\frac{1}{n}=0$$and: $$\lim_{n\rightarrow \infty }1=1,$$using the fact that the limit of the the sum of three convergent series is the sum of the limits, we see that: $$\lim_{n\rightarrow \infty }n=\lim_{n\rightarrow \infty }\left(\frac{n^2}{n+1}-\frac{1}{n}+1\right)=\lim_{n\rightarrow \infty }\left(\frac{n^2}{n+1}\right)-\lim_{n\rightarrow \infty }\left(\frac{1}{n}\right)+\lim_{n\rightarrow \infty }\left(1\right)$$$$=L-0+1=L+1,$$which is absurd since the sequence \(\{n\}_{n=1}^\infty\) is unbounded.

Answer 25

Thank you so much.