# Series problem.

1. Jan 14, 2008

### rcmango

1. The problem statement, all variables and given/known data

heres the problem: http://img527.imageshack.us/img527/9660/66702982oe9.png [Broken]

2. Relevant equations

p series or geometric series?

3. The attempt at a solution

I thought this was a p series, but i was told it was a geometric series?

anyone who can walk me through this problem, please.

Last edited by a moderator: May 3, 2017
2. Jan 14, 2008

### nicksauce

This is a geometric series. A geometric series is in the form $$\sum_n x^n$$, while a P-Series is in the form, $$\sum_n n^p$$. Ignoring the factor of 3, you can solve this series using
$$\sum_{k=101}^\infty 5^{-k} = \sum_{k=0}^\infty 5^{-k} - \sum_{k=0}^{100} 5^{-k}$$. Both right hand terms should have analytic expressions I think.

3. Jan 14, 2008

### Mathdope

In any geometric series all you need to compute the sum is the first term and the common ratio. The sum is always (first term)/(1-common ratio)

4. Jan 15, 2008

### rcmango

i know i'm close, i got 3/4 to be the sum.

heres a pic of what i've done, please correct me.

http://img518.imageshack.us/img518/3712/57667609ii9.png [Broken]

Last edited by a moderator: May 3, 2017
5. Jan 16, 2008

### Gib Z

In your original question, the numerator was not being exponentiated to the k-th power. In your evaluation it is.

It may be best if you rewrite the series as $$3 \sum_{k=101}^{\infty} \left( \frac{1}{5}\right)^k$$, and then take into account Mathdopes post.