<p>I just finished writing my Cal II exam today, and there's one problem that has been bothering me...</p>
<p>Determine if the improper integral from a=4 to b=inf of (1/(x-3)^2) converges, and if it does, where does it converge?</p>
<p>I know that it converges and I showed that (since the limit exists), but I cannot figure out WHERE it converges. Since a=4, I was able to deduce that this is basically an infinite series, where n=1 of (1/n^2). From there, I got lost.</p>
<p>I wouldn't really think of it as an infinite series because its variables change continuously. It's really just a function integrated out to infinity. I don't quite see whaere you were going with the series thing. </p>
<p>You should go about solving the problem by integrating the function just as you would if you were given a problem with rational bounds. You should get -1/(x-3) evaluated from 4 to infinity. Since -1/(infinity-3) = 1/infinity = 0 this should be trivialto evaluate.</p>
<p>I hope that helps.</p>
<p>Hmmm...so maybe I did get the answer, but misunderstood and tried to go further with the infinite series thing. I ended up with 1 as my answer (0-(-1)) for the improper integral. I'm a big choker in exams--I can't believe I didn't notice that it isn't a series since the variables change continuously and not discretely at integers...thanks for clearing that up.</p>
<p>Maybe my teacher will be nice and disregard the rest of my work.</p>