0
$\begingroup$

I was going thorugh some of my old math quizzes, and found this question, which I am currently unable to solve.

Show that the series converges. $$\sum_{n=1}^\infty a_n\,,\qquad a_n = \begin{cases}2/n & \text{if } 3\mid n \\ -1/n & \text{if } 3\nmid n\end{cases}$$

So here's what I have done.

  1. Well obviously, alternating series wouldn't work.
  2. I tried to apply the ratio test with $\limsup$, but $\limsup \left\lvert\frac{a_{n+1}}{a_n}\right\rvert = 2$, so the test was inconclusive.
  3. This series didn't look like it would converge absolutely (to me, at least), so I couldn't rearrange or group the terms and check for convergence.
  4. So I went online and found this question. Which gave me the idea to consider the $3N$-th, $3N+1$-th, $3N+2$-th partial sums respectively. Well, I could show that each of the partial sums converges, but I thought that wasn't enough because I would have to show that the 3 partial sums converge to the same value.

Should I continue on with my reasoning on 4? Or is there a better way to approach this question? Any help is greatly appreciated. Thanks in advance.

$\endgroup$
2
$\begingroup$

Let $S_n$ be the $n^\text{th}$ partial sum. You have shown $S_{3n}$, $S_{3n+1}$, and $S_{3n+2}$ each converge. Since $S_{3n}>S_{3n+1}>S_{3n+2}$, you could try to show

  • the sequence $(S_{3n})_n$ is strictly monotonically decreasing, acting as successively tighter upper bounds for all subsequent partial sums,
  • the sequence $(S_{3n+2})_n$ is strictly monotonically increasing, acting as successively tighter lower bounds for all subsequent partial sums, and
  • that $|S_{3n} - S_{3n+2}|$ converges to zero.

You might recognize that this was the method used in the proof of the alternating series test -- the partial sums alternated between an improved upper bound and an improved lower bound for all subsequent partial sums and the gap between the upper and lower bounds goes to zero.

Here, you have three sequences of partial sums -- one gives improving upper bounds, one gives improving lower bounds, and the third is always sandwiched between the other two. So the bounds are bounds for all the subsequent partial sums. If you can show the gap between the upper and lower bounds converges to zero, you will have shown that all three sequences of partial sums converge to the same thing, without having to know what that thing is.

(If you didn't have the two monotonicities, you could try to show that $| \lim_{m \rightarrow \infty, n \rightarrow \infty} (S_{3m} - S_{3n+2} )|$ converges to zero; i.e., try to show that the partial sums are enveloped by a Cauchy sequence.)

$\endgroup$
1
  • $\begingroup$ Thanks for your help! Your answer made me think that showing $\lim S_{3n} = \lim S_{3n+1} = \lim S_{3n+2}$ would be enough! $\endgroup$
    – zxcvber
    Jan 19 '20 at 12:49
1
$\begingroup$

The series converges by Dirichlet's test since the terms are of the form $a_n=b_nc_n$ with $b_n=2$ if $3\mid n$ and $b_n=-1$ otherwise, and $c_n=\frac1n$. The premises of the test are satisfied, since the partial sums of $b_n$ are bounded by $2$ and $c_n$ is monotonically decreasing and converges to $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.