# limit

• Apr 21st 2009, 10:44 PM
john_n82
limit
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 PM
Rapha
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 $n \to \infty$

$0.5* \sum_{j=0}^{\infty} \frac{1}{2^j} = 0.5 * 2 = 1$
• Apr 21st 2009, 11:23 PM
redsoxfan325
Quote:

Originally Posted by john_n82
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

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

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 $s=1$.
• Apr 22nd 2009, 12:26 AM
john_n82
Oh, so for #1, I don't have to use the definition of the limit (no epsilon)???
• Apr 22nd 2009, 01:20 AM
Rapha
Quote:

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
• Apr 22nd 2009, 08:45 AM
john_n82
are there any hints on #2, 3, 4? Thank you much.