How can I find the inverse of summation for this
$u=\sum_{h=1}^{n-j} {n-j \choose h}(-1)^h \; \frac{1}{j+h} > (1-e^{-\lambda_1X})^{\alpha \alpha_1(j+h)}$
where $x>0 , \alpha, \alpha_1 >0$
I want to find X alone in left hand side like this
$U=F(x)$ and I want to get X equals the inverse function of U,$\;\;$ $X=F^{-1}(U)$ , see the following example
$U=(1-e^{-\lambda_1X})^{\alpha \alpha_1}$
$U^{\frac{1}{\alpha \alpha_1}}=1-e^{-\lambda_1X}$
$1-U^{\frac{1}{\alpha \alpha_1}}=e^{-\lambda_1X}$
$ln(1-U^{\frac{1}{\alpha \alpha_1}})=-\lambda_1X$
$X=\frac{ln(1-U^{\frac{1}{\alpha \alpha_1}})}{-\lambda_1}$
How can I do that in my expression?
Thanks in advance