It can be proven using the Cauchy Criterion with no information about the sequence a_k, so it is true.

Suppose ∑ a_k diverges. Let ε > 0. Then there exists N such that if m ≥ n ≥ N,

|∑{k=n,m} a_k| > ε.

Combine this with the triangle equality - |∑{k=n,m} a_k| ≤ ∑{k=n,m} |a_k| - completes the proof.

Hi, as you say, it is obvious that if ∑▒a_n =∞ then ∑▒|a_n | =∞.

Perhaps you are anxious about the cases such as a_n=(-1)^(n-1), I think.

Logically, from the fact that ∑▒|a_n | : converges implies ∑▒a_n : converges,

taking contraposition, the following property is obtained:

∑▒a_n : diverges implies ∑▒|a_n | : diverges. …(*)

From the different point of view, let us check the validity of (*) by using

an ε-N argument. If ∑▒a_n : diverges,

letting S_n be the nth partial sum of the series,

∃M>0,∀n∈N,∃n_0≥∃m_0≥n, such that |S_(n_0 )-S_(m_0 ) |≥M.

The last inequality yields |a_(m_0+1)+a_(m_0+2)+⋯+a_(n_0 ) |≥M.

But the triangle inequality yields

|a_(m_0+1) |+ |a_(m_0+2) |+…+ |a_(n_0 ) |≥|a_(m_0+1)+a_(m_0+2)+⋯+a_(n_0 ) |

Combining these inequalities, we have

|a_(m_0+1) |+ |a_(m_0+2) |+…+ |a_(n_0 ) |≥M.

Thus, ∃M>0,∀n∈N,∃n_0≥∃m_0≥n, such that |〖S'〗_(n_0 )-〖S'〗_(m_0 ) |≥M,

where S_(n_ )stands for the nth partial sum of ∑▒|a_n | .

This means ∑▒|a_n | : diverges. Q.E.D.

Remark: In the case a_n=(-1)^(n-1), for a consecutive pair a_m, a_(m+1),

|a_m+ a_(m+1) | always equal to zero.

Therefore, “∀n∈N,∃n_0≥∃m_0≥n, such that |S_(n_0 )-S_(m_0 ) |≥M” is required.