# Proving that a Series Converges Conditionally

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.

• It seems to me that your sequence is $\{1, -1/2, 1/3, -1/4, ...\}$ which is different than what you wrote. – Tom Oct 13 '16 at 11:20
• @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? – user359618 Oct 13 '16 at 11:26

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.
• 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. – user359618 Oct 13 '16 at 11:30
• 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. – Adrian Oct 13 '16 at 11:34