I'm doing my homework and got stuck on these problems. Can someone please help. Thank you much.
Hi.
1) Do you have to use the hint? Alternatively there is the infinite geometric series
$\displaystyle s_n := \sum_{k=1}^n \frac{1}{2^k}$
$\displaystyle j:=k-1$
$\displaystyle s_n := \sum_{j=0}^{n-1} \frac{1}{2^{j+1}} = \sum_{j=0}^{n-1} \frac{1}{2^{j}}*2^{-1} = 1/2 * \sum_{j=0}^{n-1} \frac{1}{2^{j}}$
Let $\displaystyle n \to \infty$
$\displaystyle 0.5* \sum_{j=0}^{\infty} \frac{1}{2^j} = 0.5 * 2 = 1 $
$\displaystyle s_n-\frac{1}{2}s_n = \sum_{k=1}^n \frac{1}{2^k} - \sum_{k=1}^n \frac{1}{2^{k+1}} = \sum_{k=1}^n \frac{1}{2^{k+1}}$ $\displaystyle = \sum_{k=2}^{n+1} \frac{1}{2^k} = \sum_{k=1}^{n+1}\left(\frac{1}{2^k}\right)-\frac{1}{2} = s_{n+1}-\frac{1}{2}$
The rest of the solution to 1 lies below if you need it.
Spoiler: