1. ## sequence question

How do I interpret $\displaystyle \frac{1}{2^{(n-1)}+1}+\frac{1}{2^{(n-1)}+2}+\frac{1}{2^{(n-1)}+3}+...+\frac{1}{2^n}$

2. $\displaystyle 2^{n-1}+k$ advances from $\displaystyle k=1$ up to $\displaystyle k=2^{n-1}$, at which point $\displaystyle 2^{n-1}+2^{n-1}=2^n$

3. Originally Posted by manyarrows
How do I interpret $\displaystyle \frac{1}{2^{(n-1)}+1}+\frac{1}{2^{(n-1)}+2}+\frac{1}{2^{(n-1)}+3}+...+\frac{1}{2^n}$
The sum of the recipricals of all the natural numbers greater than $\displaystyle 2^{n-1}$ and less than or equal to $\displaystyle 2^n$

CB

4. Originally Posted by manyarrows
How do I interpret $\displaystyle \frac{1}{2^{(n-1)}+1}+\frac{1}{2^{(n-1)}+2}+\frac{1}{2^{(n-1)}+3}+...+\frac{1}{2^n}$
What do you mean by "interpret"? Were the instructions something like "describe the rule for the n-th term in words"...?

Thank you!

5. No it was to verify the inequality with that series > or = to 1/2.
How do I copy an equation from a previous thread and edit it in a new thread? Anyway , I have problem with sequences or series when the last term differs from the first. I don't know they look. For example in the above how does n=3 look compared to n=5? How do you incorporate the last term.

6. $\displaystyle \sum_{k=1}^{2^{n-1}} \frac{1}{2^{n-1} + k} \geq \sum_{k=1}^{2^{n-1}} \frac{1}{2^{n-1} + 2^{n-1}}=\frac{1}{2^n} \sum_{k=1}^{2^{n-1}} 1 = \frac{2^{n-1}}{2^n}=\frac{1}{2}.$