Convergence of a geometric series:
\(\displaystyle \sum_{n=0}^{\infty} ar^{n} \ = \ \frac{a}{1-r}, \ 0<|r|<1.\)
\(\displaystyle Hence, \ \sum_{n=2}^{\infty} 34(\frac{-1}{4})^{n} \ = \ 34[\frac{1}{16}-\frac{1}{64}+\frac{1}{256}-\frac{1}{1024}+,,,,]\)
\(\displaystyle Now \ a \ =\frac{1}{16} \ and \ r \ = \ \frac{-1}{4}, \ ergo \ Sum \ = \ 34\bigg[\frac{\frac{1}{16}}{1-(\frac{-1}{4})}\bigg]\)
\(\displaystyle = \ 34\bigg[(\frac{1}{16})(\frac{4}{5})\bigg] \ = \ \frac{17}{10}.\)
Therefore, the series converges to 17/10.
