We consider the vector space $ \mathcal F(\Re_+,\Re)=\{f:\Re_+ \to \Re\}$ of real functions defined on $\Re_+$; I want to show that the functions
$$f_n(x):\Re_+\to\Re, f_n(x)=\frac{1}{n+x},n \in \aleph_*$$
are linear independent on $ \mathcal F(\Re_+,\Re)$. What is the dimension of $ \mathcal F(\Re_+,\Re)$?
I wrote the following steps:
$$\sum_{i=1}^{n}\lambda_i f_i=\lambda_1 \frac{1}{1+x}+...\lambda_n \frac{1}{n+x}=0$$ $$\Rightarrow \lambda_1,...,\lambda_n=0$$
For $n=1$
$\lambda_1\frac{1}{1+x}=0$, with $x=0$
$\lambda_1\frac{1}{1}=0$
For $n=n+1$
$\lambda_1\frac{1}{1+x}+...+=0$, with $x=0$
$\lambda_1\frac{1}{1+x}+...+\lambda_{n+1}\frac{1}{n+1+x}=0$
I tried to isolate $\lambda_1\frac{1}{1+x}$ in order to prove that it's equal to zer0:
$(n+1+x)*(\lambda_1\frac{1}{1+x}+...)+\lambda_{n+1}=0$
Here im totally stuck, am i on the right way?
If you evaluate $$ \sum_{i=1}^{n}\lambda_i f_i(x)=\lambda_1 \frac{1}{1+x}+...+\lambda_n \frac{1}{n+x}=0 $$ at the points $x = 1, 2, 3, \ldots n$ then you get a homogeneous linear equation system for $\lambda_1,...,\lambda_n$. The coefficient determinant of this linear system is $$ \begin{vmatrix} \frac12 & \frac1{3}& \frac1{4} & \dots & \frac1{n+1} \\ \frac1{3} & \frac14 & \frac15 & \dots & \frac1{n+2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac1{n+1} & \frac1{n+2} & \frac1{n+3} & \dots & \frac1{2n} \end{vmatrix} $$ According to Find this Determinant, the value of this determinant is $$ \dfrac{\prod\limits_{1\le i<j\le n} (j-i)^2}{\prod\limits_{i,j=1}^{n}(i+j)} $$ In particular, the determinant is not zero, so that the system has only the trivial solution $\lambda_1 = ... = \lambda_n = 0$.
Hence $f_1, ..., f_n$ are linear independent.