E ewkimchi Apr 2010 57 1 May 18, 2010 #1 Consider the series Determine whether the series converges, and if it converges, determine its value. I know it converges, I got 7/10, but it's wrong. Please help! =)

Consider the series Determine whether the series converges, and if it converges, determine its value. I know it converges, I got 7/10, but it's wrong. Please help! =)

Prove It MHF Helper Aug 2008 12,883 4,999 May 18, 2010 #2 ewkimchi said: Consider the series Determine whether the series converges, and if it converges, determine its value. I know it converges, I got 7/10, but it's wrong. Please help! =) Click to expand... \(\displaystyle \sum_{n =1}^{\infty}\frac{(-7)^{n- 1}}{10^n} = \frac{1}{10} - \frac{7}{100} + \frac{49}{1000} - \frac{343}{10\,000} + \dots - \dots\). This is a geometric series with \(\displaystyle a = \frac{1}{10}\) and \(\displaystyle r = -\frac{7}{10}\). Since \(\displaystyle |r| < 1\), the series is convergent, and \(\displaystyle S_{\infty} = \frac{a}{1 - r}\) \(\displaystyle = \frac{\frac{1}{10}}{1 - \left(-\frac{7}{10}\right)}\) \(\displaystyle = \frac{\frac{1}{10}}{\frac{17}{10}}\) \(\displaystyle = \frac{1}{17}\).

ewkimchi said: Consider the series Determine whether the series converges, and if it converges, determine its value. I know it converges, I got 7/10, but it's wrong. Please help! =) Click to expand... \(\displaystyle \sum_{n =1}^{\infty}\frac{(-7)^{n- 1}}{10^n} = \frac{1}{10} - \frac{7}{100} + \frac{49}{1000} - \frac{343}{10\,000} + \dots - \dots\). This is a geometric series with \(\displaystyle a = \frac{1}{10}\) and \(\displaystyle r = -\frac{7}{10}\). Since \(\displaystyle |r| < 1\), the series is convergent, and \(\displaystyle S_{\infty} = \frac{a}{1 - r}\) \(\displaystyle = \frac{\frac{1}{10}}{1 - \left(-\frac{7}{10}\right)}\) \(\displaystyle = \frac{\frac{1}{10}}{\frac{17}{10}}\) \(\displaystyle = \frac{1}{17}\).