# Thread: the twisting and turning infinite sum of...

1. ## the twisting and turning infinite sum of...

$\displaystyle \frac{2}{5} - \frac{3}{10} + \frac{4}{15} - \frac{5}{20} + \frac{6}{25} - \frac{7}{30} + ...$

so I think the series looks like:

$\displaystyle \sum_{n=1}^{\infty}{\frac{(-1)^{n-1}*(n+1)}{5n}}$

looks a little bit geometric, the problem is to show that it either converges or diverges, we have the powerful limit comparison test at our disposal but I am not sure what to compare it to, or if there is an alternate path to solution,
I have tried the integral, basic comparison and nth term tests...
any help kindly appriciated!

thank you!!

2. the limit of the general term doesn't exist, so your series diverges.

3. the limit of the general term, the limit of n goes to infinity of $\displaystyle {\frac{(-1)^{n-1}*(n+1)}{5n}}$
I don't see how it does not exist? it looks like an indeterminate form to me, if you apply L'H we still end up with a (-1) to the infinity term in the numerator?? Should I take the log of the limit?

I hope am not missing something obvious!

4. ## Continued Confusion

So I have re-written the sum as

$\displaystyle \sum_{n=2}^{\infty}{\frac{(-1)^{n}*(n)}{5n-5}}$

But I still can't see that the limit doesn't exist, (I am aware that if it doesn't exist, then the series diverges.)

Does L'Hopital's rule work here, (does the limit comparison test work for a lower sum limit of n=2?) because I don't see any other way of showing the limit is non-existant, and showing, therefore, that the sum diverges!

Sorry if I am missing something obvious!

Thank you!

5. Originally Posted by matt.qmar
the limit of the general term, the limit of n goes to infinity of $\displaystyle {\frac{(-1)^{n-1}*(n+1)}{5n}}$
I don't see how it does not exist? it looks like an indeterminate form to me, if you apply L'H we still end up with a (-1) to the infinity term in the numerator?? Should I take the log of the limit?

I hope am not missing something obvious!
Let $\displaystyle a_n=\frac{(-1)^{n-1}(n+1)}{5n}$ (just so I don't have to keep writing it).

As $\displaystyle n\to\infty$, $\displaystyle \frac{n+1}{5n}\to\frac{1}{5}\neq0$, so the sum diverges. Conceptually, for large enough $\displaystyle n$, $\displaystyle \sum_{n>N} a_n\approx \frac{1}{5}-\frac{1}{5}+\frac{1}{5}...$

You see why that diverges?

6. thank you so much! I am not sure how I missed the (n+1)/5n.

Still not exactly sure how (-1)^infinity was taken care of, but I suppose it doesn't matter (always either +/- 1)

thanks again!

7. Originally Posted by matt.qmar
Still not exactly sure how (-1)^infinity was taken care of, but I suppose it doesn't matter (always either +/- 1)
It wasn't taken care of. The terms oscillate, and they do not approach 0; as Krizalid said, the limit does not exist.