1
$\begingroup$

So, I was suppose to prove that the series$\sum^{\infty}_{k=1}a_n$ is conditionally convergent given $\{a_n\}$ is the sequence

$$ a_n= \begin{cases} 1/k&\mathrm{if\ }n=2k-1,\\ -1/k&\mathrm{if\ }n=2k.\\ \end{cases} $$

I know that $ a_n=\{1, -1, \frac{1}{2}, -\frac{1}{2},...,\frac{1}{k}, -\frac{1}{k}, \frac{1}{k+1}, -\frac{1}{k+1}, ... \}$ but how do I show that it converges conditionally?

**correction: ** $\{1, -1/2, 1/3, -1/4, ...\}$ Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ It seems to me that your sequence is $\{1, -1/2, 1/3, -1/4, ...\}$ which is different than what you wrote. $\endgroup$ – Tom Oct 13 '16 at 11:20
  • $\begingroup$ @Tom ive got the logic of the piecewise sequence wrong. So this means that the partial sum will take the value$\frac{(-1)^{k-1}}{k}$ right and so from here I can prove conditional convergence via alternating series test and its failure to meet the criteria of absolute convergence? $\endgroup$ – user359618 Oct 13 '16 at 11:26
1
$\begingroup$

Hint:

Conditional convergence means that $\sum^{\infty}_{k=1}a_k$ converges, but $\sum^{\infty}_{k=1}\left|a_k\right|$ doesn't. Obviously,

$$\sum^{\infty}_{k=1}\left|a_k\right| = \sum^{\infty}_{k=1}\frac 1 k$$ is the harmonic series, which is divergent.

To prove the convergence of $\sum^{\infty}_{k=1}a_k$, you can use the alternating series test.

$\endgroup$
  • $\begingroup$ Hi @adjan, thanks for the hints. I have a rough sketch of which tests to use but the only problem lies with using the correct partial sum that arises from $a_n$. Im not whether my second attempt in the comments is correct tho. $\endgroup$ – user359618 Oct 13 '16 at 11:30
  • 1
    $\begingroup$ You don't need partial sums to apply the alternating series test. you only have to show that $\left|a_k\right| = \left |\frac{(-1)^k}{k}\right| = \frac{1}{k}$ decreases monotonically and converges to 0. $\endgroup$ – Adrian Oct 13 '16 at 11:34
  • $\begingroup$ 👌 thanks a lot $\endgroup$ – user359618 Oct 13 '16 at 11:38

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.