Vtyjkyuk

While revising for exams, I came across a question where at the end of it, we had to determine whether the below series was convergent or divergent:

$$sum_n=1^infty frac(-1)^nn(2+(-1)^n) $$

Unfortunately, other than knowing that as the series $ sum_n=1^infty frac(-1)^nn $ converges, this series will most likely converge as the $(2 + (-1)^n)$ terms are bounded, I can't see a way of using this to determine if it converges or not.

Here's a list of some of the other ideas I've tried, which don't seem to get me anywhere:

• 
  Generalising the series - By this, I mean replacing the $-1$ for $z$ and then try and use the ratio test to see whether $-1$ is inside or outside or on the boundary of the circle of convergence - however, when I did so, I believe that the ratio test is inconclusive for all values of $|z|$, which while disappointing, is an interesting feature of the series.


• 
  Summation by parts - While this is usually a nice tool to use for when dealing with awkward sums, I can't see any good choice of sequences $ (a_n) $ and $(b_n)$ to choose to use.


• 
  Summing consecutive terms - By this, I mean evaluating the series by looking at the sum of

$$sum_k=1^infty frac(-1)^2k2k(2+(-1)^2k) + frac(-1)^2k-1(2k-1)(2+(-1)^2k-1)$$

$$ Rightarrow sum_k=1^infty frac12k - frac13(2k-1)$$

$quad$ but then as we are left with parts of the harmonic series this doesn't seem like a nice way to $quad$evaluate it.

Other than that, I'm completely out of ideas, so any new ones (or ways to make my old ones work) would be much appreciated!

Using the following proposition will help you resolve the problem:

If $a_n longrightarrow_n to infty 0$ then $sum_n ge 1 (-1)^n a_n$ and $sum_n ge 1 (a_2n-a_2n-1)$ either both converge or both diverge.

Hint:
$$sum_n=2k-1^2k frac(-1)^nn(2+(-1)^n)=-frac12k-1+frac16k=-frac4k+16k(2k-1).$$

HINT: The Comparison Test is the gold standard for series that don't behave (like alternating series) and this leads to:

$$ frac 11+2^n ge frac 11+2^nn ge frac 11+(-2)^nnge frac 11-2^nn ge frac 11-2^n$$
Indeed, the centre series (your series rearranged) alternates between the two series either side of it, making this a very safef comparison test as the bounds are so tight.

FULL ANSWER:

Using integral tests on the leftmost and rightmost fractions worked for me, although the integration was a bit of a pain:
$$ sum_n=1^infty frac 11+2^n < int_1^infty frac 11+2^xdx +1 =frac ln(3/2)ln 2+1 \sum_n=1^infty frac 11-2^n < int_1^infty frac 11-2^xdx -1 =-2 $$
Therefore both series converge and thus your series converges (to -0.921454 with 6 decimal places.)