7

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:

$$\sum_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $$

Unfortunately, other than knowing that as the series $ \sum_{n=1}^\infty \frac{(-1)^n}{n} $ 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

$$\sum_{k=1}^\infty \frac{(-1)^{2k}}{2k(2+(-1)^{2k})} + \frac{(-1)^{2k-1}}{(2k-1)(2+(-1)^{2k-1})}$$

$$ \Rightarrow \sum_{k=1}^\infty \frac{1}{2k} - \frac{1}{3(2k-1)}$$

$\quad$ 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!

Marcelo
  • 1,382
Andrew D
  • 2,370
  • 2
    The series diverges. – Did May 26 '13 at 14:05
  • 1
    No chance of those techniques working, this series doesn't converge absolutely – Cocopuffs May 26 '13 at 14:09
  • @Did - I guess it probably doesn't help I was trying to find something to answer it incorrectly then! Is there any immediately intuitive way to see why the series would diverge, or is it just a matter of experience? – Andrew D May 26 '13 at 14:12
  • 1
    Denote $a_n=\frac{(-1)^n}{n(2+(-1)^n)}$. Note that $a_{2n-1}+a_{2n}=-\frac{1}{2n-1}+\frac{1}{6n}<-\frac{1}{3n}$. – 23rd May 26 '13 at 14:15
  • 3
    The factor $2+(-1)^n$ being alternatively $3$ and $1$ could make one suspicious that the $2k-1$ and $2k$ terms will not "cancel enough". And upon checking, their sum is about $-1/(3k)$ hence the partial sums of the series go to $-\infty$. – Did May 26 '13 at 14:15
  • @Cocopuffs Neither "absolutely" nor simply. – Did May 26 '13 at 14:16
  • By $2$'s is good. – André Nicolas May 26 '13 at 14:18
  • @Landscape - Ah, I see - so I guess I was getting there with the third idea, just that I managed to mess up the arithmetic (placing the three in the wrong fraction) and then not spot how I could bound it by the harmonic series – Andrew D May 26 '13 at 14:23
  • Denote the partial sum as $s_{n}$. If the ${s_n}$ converges, then any of its subsequence would converge to the limit. Since the subsequence ${s_{2n}}$ diverges, the original series cannot be convergent. – Nazono Sumiko Jul 07 '21 at 08:31

3 Answers3

3

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.

Marko Karbevski
  • 2,060
  • 14
  • 29
1

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

$$ \frac {1}{1+2^n} \ge \frac {1}{1+2^nn} \ge \frac {1}{1+(-2)^nn}\ge \frac {1}{1-2^nn} \ge \frac {1}{1-2^n}$$ 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: $$ \sum_{n=1}^{\infty} \frac {1}{1+2^n} < \int_1^\infty \frac {1}{1+2^x}dx +1 =\frac {\ln(3/2)}{\ln 2}+1 \\\sum_{n=1}^{\infty} \frac {1}{1-2^n} < \int_1^\infty \frac {1}{1-2^x}dx -1 =-2 $$ Therefore both series converge and thus your series converges (to -0.921454 with 6 decimal places.)

1

Hint: $$\sum_{n=2k-1}^{2k} \frac{(-1)^n}{n(2+(-1)^n)}=-\frac{1}{2k-1}+\frac{1}{6k}=-\frac{4k+1}{6k(2k-1)}.$$

Start wearing purple
  • 53,234
  • 13
  • 164
  • 223