Math Help - find the sum of the series

1. find the sum of the series

$\sum\limits_{k = 1}^\infty$ $1/(k(k+3))$

$\sum\limits_{k = 0}^\infty$ $3/ 10^k$

$\sum\limits_{k = 0}^\infty$ $1-2^k/3^k$

$\sum\limits_{k = 0}^\infty$ $2^{k+3}/3^k$

can someone tell me the general method to find the sum of the similar series, or show me an example, please!

2. $\frac{1}{k(k+3)} = \frac{1}{3} \left(\frac{1}{k}-\frac{1}{k+3}\right)$
So
$\sum_{k=1}^n \frac{1}{k(k+3)} = \frac{1}{3} \left( \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+3}\right) = \frac{1}{3} \left( \sum_{k=1}^n \frac{1}{k} - \sum_{k=4}^{n+3} \frac{1}{k}\right) =$
$=\frac{1}{3} \left( 1+\frac{1}{2}+\frac{1}{3} - \frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right)$
Therefore $\sum_{k=1}^{\infty} \frac{1}{k(k+3)} = \frac{1}{3} \left( 1+\frac{1}{2}+\frac{1}{3}\right)=\frac{11}{18}$

3. Hello, wopashui!

Sorry, there is no general method.
Each problem requires a different approach.

$(1)\;\sum_{k = 1}^{\infty} \frac{1}{k(k+3)}$
We have: . $S \;=\;\frac{1}{1\cdot4} + \frac{1}{2\cdot5} + \frac{1}{3\cdot6} + \frac{1}{4\cdot7} + \frac{1}{5\cdot8} + \cdots$

Applying Partial Fractions, we find that: . $\frac{1}{k(k+3} \;=\;\frac{1}{3}\left[\frac{1}{k} - \frac{1}{k+3}\right]$

Hence: . $S \;=\;\frac{1}{3}\bigg[\left(\frac{1}{1}-\frac{1}{4}\right) + \left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{1}{6}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \hdots \bigg]$

And everything cancels out except: . $S \;=\;\frac{1}{3}\bigg[1 + \frac{1}{2} + \frac{1}{3}\bigg]$

Therefore: . $S \;=\;\frac{1}{3}\cdot\frac{11}{6} \;=\;\frac{11}{18}$

$\sum_{k = 0}^{\infty} \frac{3}{10^k}$
$\text{We have: }\;S \;=\;\frac{3}{1} + \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \hdots \;=\;3\underbrace{\bigg[1 + \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \hdots\bigg] }_{\text{Geometric series}}$

The geometric series has the sum: . $\frac{1}{1-\frac{1}{10}} \:=\:\frac{10}{9}$

Therefore: . $S \;=\;3\cdot\frac{10}{9} \;=\;\frac{10}{3}$

$\sum_{k = 0}^{\infty} \left(1-\frac{2^k}{3^k}\right)$

$\text{We have: }\:S \;=\;\underbrace{\sum^{\infty}_{k=0}1}_{\text{dive rges}} - \underbrace{\sum^{\infty}_{k=0}\left(\tfrac{2}{3}\ right)^k}_{\text{converges}}$

Therefore, $S$ diverges.

$\sum_{k = 0}^{\infty} \frac{2^{k+3}}{3^k}$

$\text{We have: }\;S \;=\;\sum^{\infty}_{k=0}\frac{2^{k+3}}{3^k} \;=\;\sum^{\infty}_{k=0}\frac{2^3\cdot2^k}{3^k} \;=\;8\sum^{\infty}_{k=0}\left(\frac{2}{3}\right)^ k$

We have a geometric series with the sum: . $\frac{1}{1-\frac{2}{3}} \:=\:3$

Therefore: . $S \:=\:8\cdot3 \:=\:24$