Limits

mathsohard

1. Show that lim (n^n)/(n!e^n) exists without finding the limit.
n->oo

2. Show that lim (1/n){(2*4*...*(2n))/(1*3*...*(2n-1))}^2 exists without finding the limit. n->oo

I guess I have to prove this by least upper bounds, however hard to start!

dwsmith

1. Show that lim (n^n)/(n!e^n) exists without finding the limit.
n->oo

2. Show that lim (1/n){(2*4*...*(2n))/(1*3*...*(2n-1))}^2 exists without finding the limit. n->oo

I guess I have to prove this by least upper bounds, however hard to start!
For 1, the denominator grows faster than the numerator. Hence, the limit is 0.

• mathsohard

tonio

For 1, the denominator grows faster than the numerator. Hence, the limit is 0.

The same could be said about $\displaystyle \displaystyle{\frac{n}{n+1}}$ , yet this seq. does not converge to zero...

Tonio

• mathsohard

dwsmith

The same could be said about $\displaystyle \displaystyle{\frac{n}{n+1}}$ , yet this seq. does not converge to zero...

Tonio
Limits of a simple polynomial we just evaluate the fraction of the highest power variable. I would have changed my answer to take n/n then but for this case and since it doesn't say solve the limit, I gave the poster a correct answer for his limit.

• mathsohard

tonio

1. Show that lim (n^n)/(n!e^n) exists without finding the limit.
n->oo

Show that your sequence is monotone descending, which is surprisingly easy if you already know that the sequence

$\displaystyle \displaystyle{\left(1+\frac{1}{n}\right)^n$ converges monotonically ascending to $\displaystyle e$

2. Show that lim (1/n){(2*4*...*(2n))/(1*3*...*(2n-1))}^2 exists without finding the limit. n->oo

I guess I have to prove this by least upper bounds, however hard to start!

You can try here the above, but it'll be much more involved. First, be sure you can show that

$\displaystyle \displaystyle{2\cdot 4\cdot\ldots\cdot (2n)=2^n\,n!\,,\,\,1\cdot 3\cdot\ldots\cdot (2n-1)=\frac{(2n)!}{2\cdot 4\cdot\ldots\cdot (2n)}}$ , so

that now you can write your sequence as $\displaystyle \displaystye{a_n:=\frac{1}{n}\left(\frac{(2^n\,n!)^2}{(2n)!}\right)^2$ , and

now show that $\displaystyle a_{n+1}\leq a_n$ ...very carefully and slowly!

Tonio
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• mathsohard

tonio

Limits of a simple polynomial we just evaluate the fraction of the highest power variable. I would have changed my answer to take n/n then but for this case and since it doesn't say solve the limit, I gave the poster a correct answer for his limit.

I'm not trying to start an argument here but your answer isn't correct since my example contradicts it. You didn't explain anything

about polynomials, and even if you did that doesn't prove that a sequence which is given in form of a fraction will

necessarily converge to zero if simply, as you wrote, "the denominator grows faster than the numerator".

Tonio

• mathsohard

Drexel28

1. Show that lim (n^n)/(n!e^n) exists without finding the limit.
n->oo

2. Show that lim (1/n){(2*4*...*(2n))/(1*3*...*(2n-1))}^2 exists without finding the limit. n->oo

I guess I have to prove this by least upper bounds, however hard to start!

Try using the ratio test.

mathsohard

To show it is a monotone descending sequence, I just have to show that n < n+1 right?? and then how can I prove it converges???
need little more help on both of them tonio

To show it is a monotone descending sequence, I just have to show that n < n+1 right?? and then how can I prove it converges???
need little more help on both of them Well, as both sequences are obviously positive, showing they are monotonee descending automatically

makes them monotone and bounded and thus converging.

Tonio

• HallsofIvy

HallsofIvy

To show it is a monotone descending sequence, I just have to show that n < n+1 right?? and then how can I prove it converges???
need little more help on both of them To show that a sequence is monotone ascending you would need to show that $\displaystyle a_n< a_{n+1}$. To show a sequence is monotone descending, you would need to show that $\displaystyle a_n> a_{n+1}$.

If a monotone ascending sequence has an upper bound, then it converges. If a monotone descending sequence has a lower bound, then it converges. Sequences of positive numbers have, by definition of "positive", 0 as a lower bound.