Any way to compute this serie? Let $(u_n)_{n \ge1}$ such as $u_1=\frac12$ and $(\forall n \ge2):u_n=(1-\frac3{2n})u_{n-1}$. Find the value of $\displaystyle \sum_{n=1}^{\infty}u_n$. This is my try :
We can rewrite it as $\displaystyle u_n=(-1)\cdot\frac{\frac32-n}{n}\cdot u_{n-1}\Rightarrow\prod_{k=2}^nu_k=(-1)^{n-1}\cdot\prod_{n=2}^n\left(\frac{\frac32-n}n\right)\cdot\prod_{k=2}^nu_{k-1}$.
It is clear that $(\forall n \ge 0): u_n \neq 0$, thus we have $ \displaystyle u_n=(-1)^{n-1}\prod_{k=1}^n\left(\frac{\frac32-k}k\right)$.
Using the fact that $\displaystyle \binom{p}{q}=\prod_{k=1}^q \frac{p+1-k}k$ with $p=\frac12$, we now have $\displaystyle u_n=(-1)^{n-1}\binom{\frac12}{n}$.
Thus $\displaystyle \sum^\infty_{n=1}u_n=-(-1)^{-1}\binom{\frac12}{0}+\sum^\infty_{n=0}(-1)^{n-1}\binom{\frac12}{n}=1+\sum^\infty_{n=0}(-1)^{n-1}\binom{\frac12}{n}$, but here I'm stuck because i wanted to use the Newton's generalized binomial theorem yet it is not possible here....
Any help please ?
Hint (stolen from your idea of using Newton's generalized binomial theorem)
$$(1+(-1))^{1/2}=\sum_{n=0}^\infty \binom{1/2}{n}1^{1/2-n} (-1)^n=\sum_{n=0}^\infty (-1)^n\binom{1/2}{n}$$
Hint (stolen from your idea of using Newton's generalized binomial theorem)
$$(1+(-1))^{1/2}=\sum_{n=0}^\infty \binom{1/2}{n}1^{1/2-n} (-1)^n=\sum_{n=0}^\infty (-1)^n\binom{1/2}{n}$$