<p>You can't use lim n-> infinity = 0 as proof of convergence, right? If you could, that would mean 1/n converges, while we all know it diverges.</p>
<p>Asking because my book used it to show that sin(n^(1/2)) / n^(1/2) converged. If anybody could explain the proper way to do it, I'd be much obliged.</p>
<p>EDIT: I realize that's hard to read. This should make it a bit easier:</p>
<p>It seems like you're getting sequences confused with series. To show that a sequence (which in simplest terms is just a list of numbers) converges, you do take the limit as n-> infinity. If that limit exists and is finite, the sequence converges. Since it is well known that sin(x)/x -> 1 as x -> inf, the sequence {sin(n^(1/2)) / n^(1/2) } converges.</p>
<p>However, if you are dealing with a series, which is when you sum a list of terms, there are many different methods for showing convergence/divergence. The one that is encountered early on is known as the test for divergence. Here, you take the limit of the general term as n -> inf and if that limit does not exist or does not equal 0, then the series diverges. Note that this does not tell us whether the series converges, however if it is conclusive, then we know the series diverges.</p>
<p>Yep, you got that perfect. The question specified that is was a sequence, and I thought it was talking about a series. Thanks for the help.</p>
<p>I thought sin(x)/x goes to 1 as x approaches zero, not inf.</p>
<p>sin(x)/x should go to 0 as x approaches inf, right?</p>
<p>Yes, it goes to 0 as x approaches infinity.</p>
<p>It goes to 1 as x approaches 0.</p>
<p>this is a CHART with all the series/sequences tests. i found it very helpful...got me a 100 on my calc 2 exam last semester.</p>
<p>You can't use lim n-> infinity = 0 as proof of convergence, right? If you could, that would mean 1/n converges, while we all know it diverges.</p>
<p>Asking because my book used it to show that sin(n^(1/2)) / n^(1/2) converged. If anybody could explain the proper way to do it, I'd be much obliged.</p>
<h2>EDIT: I realize that's hard to read. This should make it a bit easier:</h2>
<p>I believe you are thinking of the nth term test for divergence. If the limit as n goes to infinity does not equal zero, than the series diverges. If the limit as n goes to infinity does equal zero, it is not conclusive. For 1/n, the nth term is 0, so it can't be used as a way to show convergence/divergence.</p>
<p>For sine of root n over square root n, one can use the integral to test to show that from any number to infinity, the area under the curve diverges.</p>
<p>You may also choose to think of it this way:
sin(x) is a function whose range is -1 to 1, never larger or smaller, like x is a function whose range is -inf to inf. so you can show that the denominator is growing without bound while the numerator is bounded.</p>
