consider the series
(1+1) + (0.3-.4642) + (0.03 + .2154) + (0.003 - .1000) + (.0003 + .0464) + ..........
I found the series to be:
![\Sigma_{n=0}^{\infty} {3(1/10)^n} + (-1)^{n+1}(1/\sqrt[3]{10})^n<br />](http://latex.codecogs.com/png.latex? \Sigma_{n=0}^{\infty} {3(1/10)^n} + (-1)^{n+1}(1/\sqrt[3]{10})^n<br />
)
I found it to be convergent because 3\Sigma_{n=0}^{\infty} (1/10)^n converges by geometric series test because r= |1/10| < 1
and
![<br />
\Sigma_{n=0}^{\infty} (-1)^{n+1} (1/\sqrt[3]{10})^n<br />](http://latex.codecogs.com/png.latex? <br />
\Sigma_{n=0}^{\infty} (-1)^{n+1} (1/\sqrt[3]{10})^n<br />
)
converges by geometric series test becuse the |r| < 1 also
so...
the series
math]
\Sigma_{n=0}^{\infty} {3(1/10)^n} + (-1)^{n+1}(1/\sqrt[3]{10})^n
[/tex]
would convergem but then I have to find the EXACT sum of the series and I don't know how to do this.
converges by geometric series test because r= |1/10| < 1
Originally Posted by badBKO
consider the series![\frac{3}{1-\frac{1}{10}}-\frac{1}{1+\frac{1}{\sqrt[3]{10}}}](http://latex.codecogs.com/png.latex?\frac{3}{1-\frac{1}{10}}-\frac{1}{1+\frac{1}{\sqrt[3]{10}}})
LinkBack URL
About LinkBacks