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

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- Apr 21st 2009, 10:44 PMjohn_n82limit
I'm doing my homework and got stuck on these problems. Can someone please help. Thank you much.

http://img18.imageshack.us/img18/7430/35896163.jpg - Apr 21st 2009, 11:15 PMRapha
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 $ - Apr 21st 2009, 11:23 PMredsoxfan325
$\displaystyle s_n-\frac{1}{2}s_n = \sum_{k=1}^n \frac{1}{2^k} - \sum_{k=1}^n \frac{1}{2^{k+1}} = \sum_{k=1}^n \frac{1}{2^{k+1}}$ $\displaystyle = \sum_{k=2}^{n+1} \frac{1}{2^k} = \sum_{k=1}^{n+1}\left(\frac{1}{2^k}\right)-\frac{1}{2} = s_{n+1}-\frac{1}{2}$

The rest of the solution to 1 lies below if you need it.

__Spoiler__: - Apr 22nd 2009, 12:26 AMjohn_n82
Oh, so for #1, I don't have to use the definition of the limit (no epsilon)???

- Apr 22nd 2009, 01:20 AMRapha
- Apr 22nd 2009, 08:45 AMjohn_n82
are there any hints on #2, 3, 4? Thank you much.