<p>?? thanks lin I suppose. i never said that a positive number to a power could be negative. </p>
<p>please re-read my post(s)</p>
<p>
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btw, i'm done studying, so keep shootin questions at me until I go to bed. ∞
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</p>
<p><a href="http://www.physicsforums.com/showthread.php?t=169225%5B/url%5D">http://www.physicsforums.com/showthread.php?t=169225</a>
GL.</p>
<p>haha. I meant calc AB/BC questions. stuff I can answer as a high schooler currently in calc bc.</p>
<p>basically that site you linked to is a load of giberrish.</p>
<p>I'm saying, how could e to the anything equal a negative number (even a -0)?</p>
<p>oh yea. haha whoops. good mental math on my part.</p>
<p>hmm. I really have no idea why the limit of x^2 / e^x as x approaches -∞ = -∞</p>
<p>hmm... e^-∞ = 0....... so you get ∞/0 = ∞. where DOES that damn negative come into play?</p>
<p>haha sorry for the confusion/rudeness/sarcasm. I see my mistake.</p>
<p>my bad, the answer is ∞. lol</p>
<p>Convergent or divergent? summation from 1 to infinity of (n^2-1)/(n^2+n).</p>
<p>This is probably simple, but explain, please.</p>
<p>and good job rw9700 =-).</p>
<p>Nevermind... I feel stupid, lol. I've got it...</p>
<p>grrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr. honestly. I started lookin at calc stuff to see where I went wrong.</p>
<p>oh well, at least the question is answered, and I can go to sleep peacefully tonight!</p>
<p>lin - lim n-->∞ (n^2-1)/(n^2+n) = ∞/∞</p>
<p>could do l'hopital's twice; I find it easier just to multiply the top and bottom by 1/n^2, transforming the equation to (1-(1/n^2))/(1+(1/n))</p>
<p>take the limit of that as n approaches infinity, and you get (1-0)/(1+0), which is 1. i think that validates that the series converges. I know the +n on bottom seems like it would make a difference, but in all reality, it doesn't until you go all the way up to n = infinity.</p>
<p>nope, the lim n--> ∞ has to equal 0 for it to be converging... that's the key thing I forgot about, lol. All your math is right, but your conclusion is wrong. (it's diverging, and the answer in the back of the book backs me up =-))</p>
<p>crap. I knew something was off. bah, it's so obvious now. a series doesn't converge if the sequence converges to 1........ it'd just keep adding 1 term infinitely.</p>
<p>Also (n^2-1)/(n^2+n) factors into ((n+1)(n-1))/((n)(n+1)
The (n+1) term cancels from the expression and we are left with (n-1)/(n)
We can use a direct comparison with the divergent sum 1/n to conclude that the original sum diverges =D</p>
<p>That was another way of doing the problem</p>
<p>what are all the values of p for which the infinite series sum from n=1 to infinity (n/(n^p + 1)) converges?</p>
<p>n/(n^p+1) < n/n^p = 1/n^(p-1).</p>
<p>You know that 1/n^x is convergent for x>1, so p>2. (I think.) =-)</p>
<p>Does anyone have a good general rule to use for areas rotated around the y-axis for the washer method?</p>
<p>... Uh. How does the Lagrange Error Bound work?</p>
<p>yea how do u do that, lol.</p>
<p>It basically says that the error is equal to the next term in the series.</p>
<p>So the +- error at term n is term n+1.</p>
<p>gtg to sleep =-)</p>
<p>How do you do partial fractions? I get to setting up the A/(x+a) + B/(x+b), but don't know how to go from there.</p>
<p>you need to make the partial fractions into one fraction again, so you combine them to get </p>
<p>(A(x+b) + B(x+a)) / (x+a)(x+b)</p>
<p>Then solve for A and B.</p>