# Math Help - limit

1. ## limit

I'm doing my homework and got stuck on these problems. Can someone please help. Thank you much.

2. Hi.

1) Do you have to use the hint? Alternatively there is the infinite geometric series

$s_n := \sum_{k=1}^n \frac{1}{2^k}$

$j:=k-1$

$s_n := \sum_{j=0}^{n-1} \frac{1}{2^{j+1}} = \sum_{j=0}^{n-1} \frac{1}{2^{j}}*2^{-1} = 1/2 * \sum_{j=0}^{n-1} \frac{1}{2^{j}}$

Let $n \to \infty$

$0.5* \sum_{j=0}^{\infty} \frac{1}{2^j} = 0.5 * 2 = 1$

3. Originally Posted by john_n82
I'm doing my homework and got stuck on these problems. Can someone please help. Thank you much.

$s_n-\frac{1}{2}s_n = \sum_{k=1}^n \frac{1}{2^k} - \sum_{k=1}^n \frac{1}{2^{k+1}} = \sum_{k=1}^n \frac{1}{2^{k+1}}$ $= \sum_{k=2}^{n+1} \frac{1}{2^k} = \sum_{k=1}^{n+1}\left(\frac{1}{2^k}\right)-\frac{1}{2} = s_{n+1}-\frac{1}{2}$

The rest of the solution to 1 lies below if you need it.
Spoiler:
Thus, $\frac{1}{2}s_n = s_n-\frac{1}{2}s_n = s_{n+1}-\frac{1}{2}$.

Letting $\lim_{n\to\infty} s_n = s$, we have $\frac{1}{2}s = s-\frac{1}{2}$, implying that $s=1$.

4. Oh, so for #1, I don't have to use the definition of the limit (no epsilon)???

5. Originally Posted by john_n82
Oh, so for #1, I don't have to use the definition of the limit (no epsilon)???
That's right, there is no need to use any epsilon

6. are there any hints on #2, 3, 4? Thank you much.