So, to use Alternating Sum Test, $b_n$ must be decreasing and its $\lim_{n\to \infty} = 0$.

What do we do when when one of the conditions fail? We try to take a limit/use another test?


$-\frac 25 + \frac46 - \frac67 + \frac88 - \frac {10}{9} + \ldots + $

That I think would be equal to: $\sum_{n=1}^\infty \frac{(-1)^n 2n} {4+n}$, which isn't decreasing and $\lim_{n\to \infty} b_n \neq 0$. $\lim_{n\to \infty} a_n$ then doesn't exist? Yet, it says that the sum is convergent. Where do I go wrong?


The sum is divergent by a more basic test: $$\hbox{if}\quad a_n\not\to0\quad\hbox{then}\quad\sum_{n=1}^\infty a_n\ \hbox{diverges}.$$ And in this case $$\frac{2n}{4+n}\to2$$ so $\frac{(-1)^n2n}{4+n}$ does not tend to zero.

With regard to the alternating sum test, it is a test for convergence only: you can never use it to prove that a series diverges. If the conditions of the test do not hold, the only thing you can do is to try another test.

There is no way to tell for sure which test you should use, this is why convergence of series is a difficult topic.

  • $\begingroup$ Thanks but that's not what I was asking for. My question is, how can the book say that it's convergent when $\lim b_n \neq 0$? $\endgroup$ – latenight_help Apr 30 '14 at 22:52
  • $\begingroup$ Because the book is wrong! (Or maybe you copied the question wrong? - perhaps you could double-check.) $\endgroup$ – David Apr 30 '14 at 22:53
  • $\begingroup$ No, I tried to redo it so many times, and I rechecked it at least 5 times. I guess it must be wrong. "it is a test for convergence only" makes a lot of sense now! $\endgroup$ – latenight_help Apr 30 '14 at 22:59

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