Recent content by Jimbo57

I'll be studying engineering and am thinking I won't do too much in terms of analysis, unless of course I want to. Another method in my text, which I somehow missed, was to take the ratio of A_n/A_(n+1) then simplify and take the limit as n approaches infinite. If the limit is infinite then...

Okay, that's some great knowledge that I can take forward. I'll just keep practicing on different problems and see what I get using what you showed me. Thanks!

Okay. I think I'm starting to see something, but am unsure if what I did was correct. ((n+1)n!)2/(2n+2)(2n+1)(2n) = (n!)2/(2n!) ((n+1)n!)2/(n!)2 = (2n+2)(2n+1)(2n!)/(2n!) (n+1)2 = (2n+2)(2n+1) n2+2n+1/4n2+5n+2 = 1 I see that the denominator is greater from here and this will be...

Hey Mr. Anchovy, I stumbled upon what it expands to, (2n)! = 2ⁿ*(n!)*[ 1.3.5...(2n-1) ] And see hows it's decreasing. But how (2n)! expands out to that, seems unintuitive at this point. Using An+1 = ((n+1)!)^2/(2n+2)! Which still leaves me lost. I feel like I just don't understand...

Homework Statement These trickly little buggers always seem to confuse me. I need to find out whether or not the sequence is increasing, decreasing or neither. An=(n!)2/(2n)! Homework Equations The Attempt at a Solution I'm pretty sure that it's a decreasing sequences but when I expand and...

Thank you so much LCKurtz

Gotcha, ##a_0=\frac 1 {2}^{1}## ##a_1=\frac 1 {2}^{2}## ##a_2=\frac 1 {2}^{4}## ##a_3=\frac 1 {2}^{8}## Bingo. ##a_n={2}^{2}^{n}## I can't seem to get latex to work but it's 22n EDIT: I got excited, it's 1/22n

##\frac{n}{n-1}## converges to 1 Now, dealing with the factorial is where I get lost. ##\frac{2^{n-2}}{(n-2)!}## converges to 0, but I'm not sure how.

##a_1##=##(0.5)^2## ##a_2##=##(0.25)^2## ##a_3##=##(0.0625)^2## ##a_4##=##(0.00390625)^2## Hmmm, I see that 0625 is recurring and I'm assuming that as n increases, the amount of decimal places increase by 2n places. Does that make sense? EDIT: This is probably what you didn't want me to do...

This isn't for analysis Jufro, it's Calc 101 and I have yet to see any analysis.

I'm sort of confused where the 4 came from. Is that just the result of pairing up the first 2, then showing the third 2 term as 2n-2? 4 \frac{n}{(n-1)} \frac{2^{n-2}}{(n-2)!} This is what I get after cancelling n's. But as to show how this approaches 0, I'm totally lost.

Homework Statement A bored student enters the number 0.5 in her calculator, then repeatedly computes the square of the number in the display. Taking A0 = 0.5, find a formula for the general term of the sequence {An} of the numbers that appear in the display, and find the limit of the sequence...

Hmmm are you talking about the n2 ? I don't know how else to expand the 2n/n! to cancel out n's unfortunately.

Is this generally enough to show the limit of a sequence approaches 0? I put it as my answer anyways since it seems pretty well known.

Homework Statement Find the limit of n22n/(n!) Homework Equations The Attempt at a Solution First I expand out 2n/(n!) = (2/1)(2/2)(2/3)(2/4)...(2/n) which gets increasingly small as n increases. Now, where does the n2 fit into this? I know the limit to be 0 but I can't get...