Value that this series converges to

[ladyzzyzzyzzx]

Ʃ4/(n^2-1) n=2

Need help trying to solve it. I've used other tests like the ratio test and they have all resulted inconclusive.

It does converge because the Lim as n→∞ equals 0. Do not know where to go from here. Some help would be nice! Thank you!

 

[gopher_p]

It does converge because the Lim as n→∞ equals 0.

Could you tell me the name of the test that gives this conclusion?

 

[ladyzzyzzyzzx]

Could you tell me the name of the test that gives this conclusion?

it was a test for divergence, i was told if the lim n→∞ of an ≠ 0 then it diverges

 

[gopher_p]

it was a test for divergence, i was told if the lim n→∞ of an ≠ 0 then it diverges

Does this test ever tell you that a series converges?

 

[ladyzzyzzyzzx]

Does this test ever tell you that a series converges?

not explicitly, but if it does not diverge then i am assuming it must converge

 

[gopher_p]

Well it's a common misconception, so don't feel bad. But the test for divergence is not always conclusive (in fact it often isn't). It only singles out series that don't meet the minimum requirement for convergence. It's kind of like an entrance exam; you got to pass it to get in, but it's no guarantee that you're going to be successful.

So bottom line, the test for divergence only tells you that a series diverges. It never tells you when a series doesn't diverge (i.e. it never tells you that a series is convergent).

Look at the harmonic series for a stock counterexample to the kind of reasoning you tried to use.

 

[gopher_p]

As far as your problem goes, if I tell you that you've got yourself a telescoping series, is that enough to get you going?

 

[ladyzzyzzyzzx]

As far as your problem goes, if I tell you that you've got yourself a telescoping series, is that enough to get you going?

yes actually, thank you! i just needed a direction :)

 

[gopher_p]

Good.

Now as far as having a good strategy heading into these kinds of problems (so this isn't about math, really; it's about problems you run across in math classes):

There are basically only two elementary ways that we can find the values of convergent series (remember the tests only tell you, when they're conclusive, whether or not a series converges/diverges. they don't tell you what it converges to). (1) Compute the limit of the partial sums and (2) get clever with some tricks with Power Series (don't worry if you haven't seen them). The kinds of series whose partial sums we can compute so that we can take a limit are essentially the geometric series (for which we have a formula) and telescoping series.

So if you have a problem that asks you to compute the value of a convergent series, there's a real good chance that it's either a geometric series or a telescoping series. Geometric series are relatively easy to identify, so ...

 

[Ray Vickson]

Ʃ4/(n^2-1) n=2

Need help trying to solve it. I've used other tests like the ratio test and they have all resulted inconclusive.

It does converge because the Lim as n→∞ equals 0. Do not know where to go from here. Some help would be nice! Thank you!

Your notation is unclear: is your series supposed to be
\sum \frac{4}{n^2-1} n = \sum \frac{4n}{n^2-1}, \text{ or is it }<br /> \sum \frac{4}{(n^2-1)n}?
If you mean the former, you have more-or-less written it correctly already (reading your expression according to standard priority rules), but if you mean the latter, you should use parentheses, like this: Ʃ4/[(n^2-1)n].

Assuming you mean the latter: your argument about the series converging because the nth term --> 0 is patently *false* (that is, the argument is false, not necessarily the conclusion). Lots of divergent series have terms that --> 0; it is just that they do not --> 0 fast enough. That is the issue you must decide.

RGV