1. ## Telescoping Series

Can somebody help with this?

Evaluate as a telescoping series: The summation from k=1 to infinity of (k-1)/2^(k+1).

The answer is 1/2 but I need to see the workings so I can figure out how it was arrived at.

2. Originally Posted by p75213
Can somebody help with this?

Evaluate as a telescoping series: The summation from k=1 to infinity of (k-1)/2^(k+1).

The answer is 1/2 but I need to see the workings so I can figure out how it was arrived at.
Hi

$\displaystyle \frac{k-1}{2^{k+1}} = \frac{k}{2^k}-\frac{k+1}{2^{k+1}}$

3. Originally Posted by running-gag
Hi

$\displaystyle \frac{k-1}{2^{k+1}} = \frac{k}{2^k}-\frac{k+1}{2^{k+1}}$
Would I be right in saying you should also show that $\displaystyle \lim_{k \to \infty} \frac{k-1}{2^k} = 0$ (trivial really) since otherwise the telescoping thing might not work...

Would I be right in saying you should also show that $\displaystyle \lim_{k \to \infty} \frac{k-1}{2^k} = 0$ (trivial really) since otherwise the telescoping thing might not work...

Since $\displaystyle \frac{k-1}{2^{k+1}} = \frac{k}{2^k}-\frac{k+1}{2^{k+1}}$

$\displaystyle \sum_{k=1}^{N} \frac{k-1}{2^{k+1}} = \sum_{k=1}^{N} \left(\frac{k}{2^k}-\frac{k+1}{2^{k+1}}\right) = \frac12 - \frac{N+1}{2^{N+1}}$

which converges towards $\displaystyle \frac12$ as N approaches infinity

5. Originally Posted by running-gag

Since $\displaystyle \frac{k-1}{2^{k+1}} = \frac{k}{2^k}-\frac{k+1}{2^{k+1}}$

$\displaystyle \sum_{k=1}^{N} \frac{k-1}{2^{k+1}} = \sum_{k=1}^{N} \left(\frac{k}{2^k}-\frac{k+1}{2^{k+1}}\right) = \frac12 - \frac{N+1}{2^{N+1}}$

which converges towards $\displaystyle \frac12$ as N approaches infinity
Lol yeah I know. I've been reading about the whole telescoping sum thing recently and it can easily be wrongly applied to infinite sums to show things like 0=1.

Was just being annoyingly rigorous and hoping OP would catch on and realize either you must show it $\displaystyle \to 0$ or do it your way with only summing to N terms.

6. You are right to be rigorous !

7. Hi Guys,
This is the part I need explained:-

(k-1)/2^(k+1) = k/2^k-k+1/2^(k+1).

8. $\displaystyle \frac{k}{2^k} = \frac{2k}{2^{k+1}}$ (multiplying numerator and denominator by 2)

$\displaystyle \frac{k}{2^k}-\frac{k+1}{2^{k+1}} = \frac{2k}{2^{k+1}}-\frac{k+1}{2^{k+1}} = \frac{2k-(k+1)}{2^{k+1}} = \frac{k-1}{2^{k+1}}$

9. How would we handle k-1/3^(k+1)? 2k/3^(K+1) won't break down into k/3^k.

If we try the following: 3k/3^(k+1) - 2k+1/3^(k+1) = k/3^k - 2k+1/3^(k+1). Which is no longer a telescoping series ie. the middle bits don't cancel.

10. $\displaystyle \frac{k-1}{3^{k+1}} = \frac{2k-1}{4 \cdot 3^k} - \frac{2(k+1)-1}{4 \cdot 3^{k+1}}$