Find the sum of the series

[amj]

Find the sum of the series.



I'm pretty sure this is a geometric series, but I'm having trouble rewriting it in the proper form. Thanks for the help.

 

[tkhunny]

If that is a Geometric Series, please state the Common Ratio.

 

[ksdhart2]

Find the sum of the series.

\(\displaystyle \displaystyle \sum_{n=0}^{\infty} \dfrac{2^{n-1}-1}{7^n}\)

I'm pretty sure this is a geometric series, but I'm having trouble rewriting it in the proper form. Thanks for the help.

I'd begin by dividing it into two sums:

\(\displaystyle \displaystyle \sum_{n=0}^{\infty} \dfrac{2^{n-1}-1}{7^n} = \sum_{n=0}^{\infty} \dfrac{2^{n-1}}{7^n} - \sum_{n=0}^{\infty} \dfrac{1}{7^n}\)

Then pull out a 7 from the denominator of the first series:

\(\displaystyle \displaystyle = \dfrac{1}{7} \cdot \sum_{n=0}^{\infty} \dfrac{2^{n-1}}{7^{n-1}} - \sum_{n=0}^{\infty} \dfrac{1}{7^n}\)

Can you finish up from here?

 

[tkhunny]

Aw, you gave it away...

 

[amj]

But since the lower bound of the first series is n = 0, doesn't that mean that r cannot be raised to n-1, and that it should instead be raised to n?

 

[tkhunny]

It is of no consequence. |r|<1 You do the arithmetic and see what you get.