Alternating Series - Converge or Diverge?

[Jason76]

\(\displaystyle \sum_{n = 1}^{\infty} \dfrac{(-1)^{n}4}{n}\)

\(\displaystyle a_{n + 1} = \dfrac{4}{n + 1}\)

\(\displaystyle a_{n} = \dfrac{4}{n}\)

\(\displaystyle n = 5\) and \(\displaystyle a_{n + 1} = \dfrac{4}{6}\) and \(\displaystyle a_{n} = \dfrac{4}{5}\)

How did they come up with the value of n?

Alternating Series Test

If \(\displaystyle 0 < a_{n + 1} \leq a_{n}\) and \(\displaystyle \lim n \rightarrow \infty a_{n} = 0\) then it converges, otherwise diverges.

Ok, this last bit makes sense.

So \(\displaystyle 0 < \dfrac{4}{6} \leq \dfrac{4}{5}\) and \(\displaystyle \lim n \rightarrow \infty \dfrac{4}{n} = 0\) so this alternating series converges.

 

[Jason76]

they reached into the sky and chose one as an example....

Oh, I see, since it's a series, they could choose any number (or maybe, any positive number). Is this correct?

 

[Jason76]

OK, makes sense.

 

[Jason76]

Right but wouldn't it be better to choose n =1, the first in the series?

In that case you would have \(\displaystyle 0 < -4 < 2\) which would also indicate that it converges (assuming the \(\displaystyle \lim n \rightarrow \infty = 0\)

 

[Jason76]

But, if \(\displaystyle 5\) is chosen as the term, then wouldn't this be the case?

\(\displaystyle \dfrac{(-1)^{5}(4)}{5} = \dfrac{-20}{5} = -4\)

\(\displaystyle \dfrac{(-1)^{5 + 1}(4)}{5 + 1} = \dfrac{(-1)^{6}(4)}{6}\dfrac{24}{6} = 4\)

\(\displaystyle 0 < 4 < -4\) which is NOT true.

or should the \(\displaystyle (-1)\) part be ignored?

 

[Jason76]

So, I'm assuming the n exponent of \(\displaystyle (-1)\) is also ignored.

\(\displaystyle n = 5\)

\(\displaystyle \dfrac{(4)}{5} = \dfrac{4}{5} \)

\(\displaystyle \dfrac{(4)}{5 + 1} = \dfrac{(4)}{6} = \dfrac{2}{3}\)

\(\displaystyle 0 < \dfrac{2}{3} < \dfrac{4}{5}\) - converge

\(\displaystyle n = 1\)

\(\displaystyle \dfrac{(4)}{1} = \dfrac{4}{1} = 4\)

\(\displaystyle \dfrac{(4)}{1 + 1} = \dfrac{(4)}{2} = 2\)

\(\displaystyle 0 < 2 < 4\) - converge

 

[lookagain]

But, if \(\displaystyle 5\) is chosen as the term, then wouldn't this be the case?

\(\displaystyle \dfrac{(-1)^{5}(4)}{5} = \dfrac{-20}{5} = -4 \ \ \ \ \)No, that equals \(\displaystyle \ \frac{-4}{5}.\)

\(\displaystyle \dfrac{(-1)^{5 + 1}(4)}{5 + 1} = \dfrac{(-1)^{6}(4)}{6}\dfrac{24}{6} = 4 \ \ \ \ \)No, that equals \(\displaystyle \ \frac{2}{3}.\)

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