Evaluate $\sum_{k=1}^\infty \frac{k-1}{2^{k+1}}$ as a telescoping series? Okay, so my book shows (with absolutely no explanation) an equivalency to 
$\lim\limits_{x \to \infty} (\frac{1}{2} - \frac{x+2}{2^x})$
, and I have no idea where this came from.  From my understanding, telescoping series generally telescope down as a result of combining positives and negatives and may lead to a result of the first term minus the last term in the series ($\infty$ in such as case).  I do see a pattern in the sequence once I split into 
$\frac{k}{2^{k+1}}$ and $\frac{-1}{2^{k+1}}$ but I'm not sure of there being any telescoping in the way the lesson examples show.  
-Can someone tell me how $\lim\limits_{x \to \infty} (\frac{1}{2} - \frac{x+2}{2^x})$ was obtained, since I have no explanation in the chapter or solution?
 A: $a_k$ is a telescopic term if $a_k$ can be written as $b_{k}-b_{k+1}$. If $a_k=\frac{k-1}{2^k}$, an we find some simple sequence $\{b_k\}_{k\geq 1}$ with such property? It looks reasonable. We may try $b_k=\frac{k}{2^k}$, for instance.
$$ b_k-b_{k+1}= \frac{k}{2^k}-\frac{k+1}{2^{k+1}} = \frac{k-1}{2^{k+1}}=a_k $$
It works, so
$$ \sum_{n=1}^{N}a_n = \sum_{n=1}^{N}(b_{n}-b_{n+1})=b_1-b_{N+1}=\frac{1}{2}-\frac{N+1}{2^{N+1}} $$
and the limit of the RHS as $N\to +\infty$ clearly equals $\frac{1}{2}$.
A: Given the formula you have, the thing to try would be
$$
\frac{k}{2^k}-\frac{k+1}{2^{k+1}}=\frac{k-1}{2^{k+1}}\tag1
$$
Now the sum can be written as
$$
\sum_{k=1}^n\frac{k-1}{2^{k+1}}=\sum_{k=1}^n\left(\frac{k}{2^k}-\frac{k+1}{2^{k+1}}\right)\tag2
$$
Note that this does not give the same partial sum as given in the question.
A: Apparently there was a typo somewhere: “$\frac{x + 2}{2^x}$” should have being replaced by “$\frac{x}{2^x}$”. That being said, let $s_x \mathop{:=} 1/2 - x \mathbin/ 2^x$ for $x \in \mathbf{N}^*$. Observe that
$$
s_{x + 1} - s_x = \frac{1}{2} - \frac{x + 1}{2^{x + 1}} - \biggl(\frac{1}{2} - \frac{x}{2^x}\biggr) = \frac{x}{2^x} - \frac{x + 1}{2^{x + 1}} = \frac{2 x}{2 \cdot 2^x} - \frac{x + 1}{2 \cdot 2^x} = \frac{2x - (x + 1)}{2^{x + 1}}
\text,$$
which is finally $(x - 1) \mathbin/ 2^{x + 1}$. Let us denote that value $u_x$; then the initial problem is to compute
$$
\sum_{k = 1}^{\infty} u_k = \sum_{k = 1}^{\infty} (-s_k + s_{k + 1})
\text.$$
If we consider the truncated sum for $1 \leq k \leq x - 1$, we get
$$
(-s_1 + s_2) + (-s_2 + s_3) + \cdots + (-s_{x - 2} + s_{x - 1}) + (-s_{x - 1} + s_x)
\text,$$
where almost all the terms “telescope” by appearing once positively and once negatively, finally yielding
$$
\sum_{k = 1}^x u_k = -s_1 + s_x
\text.$$
As $s_1 = 0$, this further simplifies into “$s_x$”; so that finally, letting $x$ tend to infinity, the sum of the infinite series is the limit of $s_x$ when $x \to \infty$, namely $1/2$.
