linear independence of functions I am trying to prove that the set $\left\lbrace\frac{1}{n+x}\right\rbrace_{n \in \mathbb{N}}$ is linear independant in the Vector space of functions from $\mathbb{R}_{>0}$ $\to$ $\mathbb{R}$.
So starting with the linear combination $\lambda_1\frac{1}{1+x} +\cdots +\lambda_{k+1}\frac{1}{k+1+x}$
,the idea was to convert it to a common denominator. Thus the polynomial in the counter is of degree k, having a maximum of k roots. This implies that the polynomial in the counter is the zero polynomial.
My problem so far is, that the only information i have about the system of equations is the first one  $\lambda_1+\cdots +\lambda_{k+1}=0$.
Does anyone know how to show that all lambdas are zero?
 A: Notice that the sum needs to be the zero constant function. So:
$$ \lambda_1 \frac{1}{1+x}+\cdots + \lambda_n \frac{1}{n+x} = 0 \Rightarrow \sum_{i=1}^n \lambda_i\prod_{j\neq i}(j+x) = 0 $$
Taking $x = -i$, we have
$$  \lambda_i\prod_{j\neq i}(j-i) = 0 \Rightarrow \lambda_i = 0  $$
since $\prod_{j\neq i}(j-i) \neq 0$.
A: The linear independence can be investigated through the Wronskian
$$
W_n(x)=\begin{vmatrix}
+\frac{1}{1+x}     & +\frac{1}{2+x}     & \ldots & +\frac{1}{n+x}     \\
-\frac{1}{(1+x)^2} & -\frac{1}{(2+x)^2} & \ldots & -\frac{1}{(n+x)^2} \\
\ldots             & \ldots             & \ldots & \ldots             \\
\frac{(-1)^{n+1}}{(1+x)^n} & \frac{(-1)^{n+1}}{(2+x)^n} & \ldots & \frac{(-1)^{n+1}}{(n+x)^n}
\end{vmatrix}
$$
The calculation of the Wronskian seems to be (by various tries with Wolfram Mathematica)
$$
W_n(x)=\frac{a_n}{(1+x)^n(2+x)^n\cdot\ldots\cdot(n+x)^n}
$$
where $a_n$ is the OIES Sequence A055209, so it is different from zero and the functions are then independent.
I leave to someone clever than me the proof that the Wronskian is effectively given by the previous expression.
