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Maths question

Will these kind of questions ever come up in STEP? What level is this?

Show that limn(1)n+1n \lim _{n\rightarrow \infty } (-1)^{n} + \dfrac{1}{n} does not exist.
Show that limnsin(n) \lim _{n\rightarrow \infty } sin(n) does not exist.

Thanks
(edited 7 years ago)
Original post by Concav
Will these kind of questions ever come up in STEP? What level is this?

Show that limn(1)n+1n \lim _{n\rightarrow \infty } (-1)^{n} + \dfrac{1}{n} does not exist.
Show that limnsin(n) \lim _{n\rightarrow \infty } sin(n) does not exist.

Thanks


Unless it's like explain, using a graph, why the limit doesn't exist, actually proving it would require knowledge of the definition of a limit which is not A Level material. So I don't think they'd require an actual proof using the definition of the limit.

I've tagged @Zacken, as he'll know more about STEP than I do.
Reply 2
Original post by Concav
Will these kind of questions ever come up in STEP? What level is this?

Show that limn(1)n+1n \lim _{n\rightarrow \infty } (-1)^{n} + \dfrac{1}{n} does not exist.
Show that limnsin(n) \lim _{n\rightarrow \infty } sin(n) does not exist.

Thanks


Agree with the above, wouldn't come up in STEP apart from a quick tiny part of a question that asks you to explain 'obviously' why the limit doesn't exist or along those lines; wouldn't bother with it, though.
Original post by Concav
Will these kind of questions ever come up in STEP? What level is this?

Show that limn(1)n+1n \lim _{n\rightarrow \infty } (-1)^{n} + \dfrac{1}{n} does not exist.
Show that limnsin(n) \lim _{n\rightarrow \infty } sin(n) does not exist.

Thanks


Agree with @Zacken and add that this is exactly the sort of question that you get in first year university analysis examples sheets.
Reply 4
Original post by Gregorius
Agree with @Zacken and add that this is exactly the sort of question that you get in first year university analysis examples sheets.


http://imgur.com/hhCrmPw Is that right for the first part of the question?
Reply 5
Original post by Concav
http://imgur.com/hhCrmPw Is that right for the first part of the question?


A few remarks.

1. A monotonically increasing sequence (an)(a_n) means that an+1ana_{n+1} \geq a_n. Period. It does not mean that an+1an1\frac{a_{n+1}}{a_n} \geq 1.

Some counterexamples:

(i) (an)=1,1,2,3,4,(a_n) = -1, 1, 2, 3, 4, \cdots is a monotonically increasing sequence but a2a1=11\frac{a_2}{a_1} = -1 \leq 1.

(ii) (an)=0,1,2,(a_n) = 0, 1, 2, \cdots is a monotonically increasing sequence but a2a1=10=???\frac{a_2}{a_1} = \frac{1}{0} = ???.

2. Next, you say that n0n\geq 0 but your function isn't defined for n=0n=0...

3. Third, you've proved the wrong direction. You've assumed an+1>ana_{n+1} > a_n and gotten to a true statement without making sure you use double implication arrows. You need to start with n+1n1\frac{n+1}{n} \geq 1 and then get to an+1>ana_{n+1} > a_n
Reply 6
Original post by Zacken
A few remarks.

1. A monotonically increasing sequence (an)(a_n) means that an+1ana_{n+1} \geq a_n. Period. It does not mean that an+1an1\frac{a_{n+1}}{a_n} \geq 1.

Some counterexamples:

(i) (an)=1,1,2,3,4,(a_n) = -1, 1, 2, 3, 4, \cdots is a monotonically increasing sequence but a2a1=11\frac{a_2}{a_1} = -1 \leq 1.

(ii) (an)=0,1,2,(a_n) = 0, 1, 2, \cdots is a monotonically increasing sequence but a2a1=10=???\frac{a_2}{a_1} = \frac{1}{0} = ???.

2. Next, you say that n0n\geq 0 but your function isn't defined for n=0n=0...

3. Third, you've proved the wrong direction. You've assumed an+1>ana_{n+1} > a_n and gotten to a true statement without making sure you use double implication arrows. You need to start with n+1n1\frac{n+1}{n} \geq 1 and then get to an+1>ana_{n+1} > a_n


Thanks for the help :smile:

I understand i assumed it but could you start me off how to commence with the proof?
Reply 7
Original post by Concav
Thanks for the help :smile:

I understand i assumed it but could you start me off how to commence with the proof?


Start with (n+1)n>nn(n+1)^n > n^n
Reply 8
Original post by Zacken
Start with (n+1)n>nn(n+1)^n > n^n


Of course that is true especially if n is non negative. Just dont see how to go from that to the original sequence.
Reply 9
Original post by Concav
Of course that is true especially if n is non negative. Just dont see how to go from that to the original sequence.


Multiply both sides by nn\frac{n}{n}, think about how you can get factorials from that. Spend a good while thinking about it.

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