# Additional properties of finite series that have the property: $\sum a_k = -\sum \frac{a_k}{k}$

Dear Math Stack Exchange Community,

I have a generator that is producing sets of integer, non-zero terms $\{a_1, a_2, a_3, ..., a_n\}$ such that these terms have the following property:

$$\sum \limits_{k=1}^{n}a_k = -\sum\limits_{k=1}^{n}\frac{a_k}{k}$$

I was wondering if all series that have these properties fall under a class of series that might have additional interesting properties?

While all sets of these terms have the above property, some also have the property where they sum to zero. Is there a more specific class or any other interesting properties when they sum to zero?

Thanks everyone who reads this for taking the time to look at it! Any help is greatly appreciated.

• there is a minus sign in the headline but not in the text. – Andreas May 31 '18 at 16:48
• The term on the right is not defined for k=0. – Martin R May 31 '18 at 16:51
• Thanks guys! Made the edits. – user122523 May 31 '18 at 16:58

let $v_n = '(-1,-1/2,\ldots,-1/n)$ and $e_n = (1,1,\ldots,1)$ what you are asking for is for which $x_n=(a_1,\ldots,a_n)$
$$x_n\cdot v_n = x_n \cdot e_n = c_0$$
each of those conditions represents an affine $(n-1)$ dimensional hyper space that is orthogonal and passes through a point on the line that the vector e.g. $v_n$ span. $v_n,e_n$ is linerarly independent so the space of solutions $x_n$ is a $n-2$ dimensional space. let $u_1,\ldots u_{n-2}$ span this space, Then $x_n$ can be written $$x_n = b_1u_1 + \cdots + b_{n-2}u_{n-2}+\alpha e_n + \beta v_n$$ Now the condition can be written,
$$c_0 = x_n\cdot e_n = \alpha e_n \cdot e_n + \beta v_n \cdot e_n = \alpha n + \beta A_n$$ $$c_0 = x_n\cdot v_n = \alpha e_n \cdot v_n + \beta v_n \cdot v_n = \alpha A_n + \beta B_n$$ with, $$A_n = \sum_{i=1}^n - 1/k$$
$$B_n = \sum_{i=1}^n 1/k^2$$
Therefore a solution will be given by solving $$\begin{pmatrix} n & A_n \newline A_n & B_n \end{pmatrix} \begin{pmatrix} \alpha \newline \beta \end{pmatrix} = \begin{pmatrix} c_0 \newline c_0 \end{pmatrix}$$ Denote the matrix by $A$, then this is solvable if, $$\det(A) = n B_n - A_n^2 = (n \sum \frac{1}{k^2}) - \left ( \sum \frac{1}{k} \right )^2\neq 0$$ Assuming this we can find a solution $\alpha,\beta$ for all $c_0$. Especially if $c_0= 0$ then $\alpha=0,\beta=0$ if we assume the determinant is nonzero.