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.
I was just giving a hint to start with
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.