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# Analysis questions (sequences) watch

1. https://www8.imperial.ac.uk/content/...1P1/m1p1s4.pdf

I need help on questions 5 and 6.

For 5, how would I prove the bit given for the hint? Induction won't help.

For 6, I tried to sandwich it between two sequences but it didn't help.
2. 5. Obviously n^(1/n) is always greater than 1. (Why?) Then if you show that, given any r, n^(1/n) < r for sufficiently large n, you're done. (Why?) So, to do that... well, you know that n/(r^n) --> 0 as n --> infinity, so in particular, given fixed r, we have n/(r^n) < 1 for sufficiently large n. Can you manipulate this to the inequality you want?

6. Write out the first ten(ish) terms of the sequence for M = 4 or 5, see what happens, just to get a feel for it. When you've done that, you might see why this (admittedly rather big, hence split into two parts) hint is helpful:

Spoiler:
Show
The sequence starts decreasing when n > M, so let's just look at the sequence from there. Let K = M^M/M! (the Mth term of the sequence), which is constant. Then, write a few terms of your sequence (starting from the Mth term - ignore the start, it won't affect convergence) in terms of K and M.
Spoiler:
Show
K, KM/(M+1), KM^2 / (M+1)(M+2), ... - why does this converge to 0?

Remember for question 6 that you don't need to sandwich it between two sequences - using 0 as a lower bound is good enough, so you just need to find an upper bound tending to 0. Don't make your job harder than it already is!
3. (Original post by generalebriety)
5. Obviously n^(1/n) is always greater than 1. (Why?) Then if you show that, given any r, n^(1/n) < r for sufficiently large n, you're done. (Why?) So, to do that... well, you know that n/(r^n) --> 0 as n --> infinity, so in particular, given fixed r, we have n/(r^n) < 1 for sufficiently large n. Can you manipulate this to the inequality you want?

6. Write out the first ten(ish) terms of the sequence for M = 4 or 5, see what happens, just to get a feel for it. When you've done that, you might see why this (admittedly rather big, hence split into two parts) hint is helpful:

Spoiler:
Show
The sequence starts decreasing when n > M, so let's just look at the sequence from there. Let K = M^M/M! (the Mth term of the sequence), which is constant. Then, write a few terms of your sequence (starting from the Mth term - ignore the start, it won't affect convergence) in terms of K and M.
Spoiler:
Show
K, KM/(M+1), KM^2 / (M+1)(M+2), ... - why does this converge to 0?

Remember for question 6 that you don't need to sandwich it between two sequences - using 0 as a lower bound is good enough, so you just need to find an upper bound tending to 0. Don't make your job harder than it already is!

5. I managed to manipulate it to that. But I still don't see why it completes the proof.

For 6.
Someone told me that they simply used the ratio test to prove that it converges to 0.

To show that the sequence is unbounded they showed me this:

M^n < n! (because it goes to zero so the bottom of the fraction must be larger)

M < (n!)^(1/n)

So we can make (n!)^(1/n) larger than M and M can be any number. So it is unbounded.
4. (Original post by gangsta316)
5. I managed to manipulate it to that. But I still don't see why it completes the proof.
Huh? I don't understand what you've done and what you haven't done.

(Original post by gangsta316)
So we can make (n!)^(1/n) larger than M
Why?
5. Is this assessed work, by the way?
6. (Original post by generalebriety)
Huh? I don't understand what you've done and what you haven't done.

Why?

For 5:

I've shown that, for n large enough, n^(1/n) < r. But I don't see how this proves that n^(1/n) has limit 1. If we accept that n^(1/n) is greater than 1 for all n (perhaps prove by induction or something) then we can say

|n^(1/n) - 1| < r - 1

And we can let epsilon = r - 1 for epsilon between 0 and 1.

For 6:

M^n < n! (because it goes to zero so the bottom of the fraction must be larger)

M < (n!)^(1/n)

So we can make (n!)^(1/n) larger than M and M can be any number. So it is unbounded. Is that ok?

(Original post by DFranklin)
Is this assessed work, by the way?
No.
7. (Original post by gangsta316)
I don't see how this proves that n^(1/n) has limit 1.
That wasn't what the question was asking you to do, was it?

(Original post by gangsta316)
M^n < n! (because it goes to zero so the bottom of the fraction must be larger)
I'm sorry, I think I must be misunderstanding, but isn't this what you're trying to prove all along?
8. (Original post by gangsta316)
For 5:

I've shown that, for n large enough, n^(1/n) < r. But I don't see how this proves that n^(1/n) has limit 1. If we accept that n^(1/n) is greater than 1 for all n (perhaps prove by induction or something) then we can say

|n^(1/n) - 1| < r - 1

And we can let epsilon = r - 1 for epsilon between 0 and 1.
Since n > 1 it follows that n^(1/n) > 1; you don't need induction here. The rest of your thoughts are along the right lines, but they're not really a proof. I would want to see something that starts: "Take , we will show that for n sufficiently large, "

For 6:

M^n < n! (because it goes to zero so the bottom of the fraction must be larger)

M < (n!)^(1/n)

So we can make (n!)^(1/n) larger than M and M can be any number. So it is unbounded. Is that ok?
Again the idea is fine, but you need to put your thoughts in the right order. I would start: "Suppose (for contradiction) that (n!)^(1/n) is bounded by M. We know that ..."
9. @GE: Both questions have two parts I think (though I'm speaking from memory, now).
10. (Original post by DFranklin)
@GE: Both questions have two parts I think (though I'm speaking from memory, now).
Oh. I got confused by the fact that the OP didn't seem to be following my hints. Perhaps it was my fault all along.
11. (Original post by generalebriety)
Oh. I got confused by the fact that the OP didn't seem to be following my hints. Perhaps it was my fault all along.
I think someone else (at his college) basically gave him another method that he was using instead.
12. I think I managed to do both of these in a tutorial today. I had basically done them correctly.

6:
You can do the first bit by the ratio test.

For the second bit. Fix M > 0. There exists N such that n > N implies that (this N exists because the sequence goes to zero so we can set epsilon to something less than 1 and find N)

M^n/(n!) < 1

M^n < n!

M < (n!)^(1/n)

And we're done. It's unbounded.

And for 5 if we show that n large enough implies that n^(1/n) < r then

|n^(1/n) - 1 | < r - 1

Now set epsilon = r - 1. The above is true for all r > 1 so it is true for all epsilon > 0. So we're done.

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Updated: February 10, 2010
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