explicit formula for the series For each series, find an explicit formula for the sequence of partial sums and determine if the series converges. 
(a)  $\sum_{n=1}^{∞} 1/2^{n}$ 
(b) $\sum_{n=1}^{∞} 1/(n(n + 1))$ 
(c) $\sum_{n=1}^{∞} \log ({(n +1)/n})$ 
(In (c), log(x) refers to the natural logarithm function from calculus.)
I know that part (a) is a geometric series and (b) and (c) are telescoping series but I am unsure of how to find an explicit formula.
 A: HINT $1$:
$$\sum_{k=M}^Nr^k=\frac{r^M-r^{N+1}}{1-r}\,\,\,r\ne 1$$
HINT $2$:
$$\sum_{k=M}^N\left(a_{n+1}-a_{n}\right)=a_{N+1}-a_{M}$$

SPOILER ALERT Scroll over the highlighted area to reveal the solution

 For the geometric sum $S(M,N)=\sum_{k=M}^Nr^k$, we have that $$\begin{align}rS(M,N)-S(M,N)&=\sum_{k=M}^Nr^{k+1}-\sum_{k=M}^Nr^{k}\\\\&=r^{N+1}-r^M\\\\S(M,N)(r-1)&=r^{N+1}-r^M\\\\S(M,N)&=\frac{r^M-r^{N+1}}{1-r}\end{align}$$If $M=1$, $r=1/2$, and $N\to \infty$, we have $$\sum_{k=1}^\infty \frac{1}{2^k}=\frac{1/2}{1-(1/2)}=1$$

A: Yes, you are right, the first part is geometric with a convergent sum. I am quite comfortable writing it in the form $\sum_{n=1}^\infty a_0r^{n-1}$. You can rewrite it like this: $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{2}\left (\dfrac{1}{2} \right)^{n-1} = \dfrac{1/2}{1-(1/2)}. $ (why?)
Second and third items are telescoping. You can rewrite $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$. List down the first few terms and discover that terms cancel each other out (hence telescoping). As $n \to \infty$, does the sum converge? 
Recall your logarithm properties: $\log(a/b) = \log a - \log b$. So this is, $\log (n+1) - \log(n)$. So it is still telescoping. But, is it convergent? 
