# Sum of series

Hi, I'm having trouble getting the sum of the following series. I'm pretty sure that it is convergent.

$$\sum\limits_{n = 1}^\infty {\frac{{\left( { - 3} \right)^n }}{{7^n }}}$$

Here is what I have done.

$$\sum\limits_{n = 1}^\infty {\frac{{\left( { - 3} \right)^n }}{{7^n }}} = - \frac{3}{7} + \frac{9}{{49}} - \frac{{27}}{{343}} + ...$$

$$s_1 = - \frac{3}{7},s_2 = - \frac{{12}}{{49}},s_3 = - \frac{{111}}{{343}}$$

The sum seems to be heading towards a negative number.

I think I need to deduce a general form for s_n and then take the limit as n goes to infinity. I'm having trouble with this. I've only been able to come up with:

$$s_n = \frac{{\left( { - 1} \right)\left( {something + ve} \right)}}{{7^n }}$$

I've decided to take a slightly different approach to this.

$$s_n = \frac{{7\left( {7^{n - 1} s_{n - 1} } \right) - 3^n }}{{7^n}} = s_{n - 1} - \left( {\frac{3}{7}} \right)^n$$

$$s_n = s_{n - 2} - \left( {\frac{3}{7}} \right)^{n - 1} - \left( {\frac{3}{7}} \right)^n$$

$$s_n = s_0 - \sum\limits_{k = 1}^n {\left( {\frac{3}{7}} \right)} ^k$$

$$\sum\limits_{n = 1}^\infty {\frac{{\left( { - 3} \right)^n }}{{7^n }}} = \mathop {\lim }\limits_{n \to \infty } s_n = - \frac{3}{7} - \sum\limits_{k = 1}^\infty {\left( {\frac{3}{7}} \right)} ^k = - \frac{3}{7} - \left( {\frac{{\frac{3}{7}}}{{1 - \frac{3}{7}}}} \right) = - \frac{{33}}{{28}}$$

Ok well that's my clumsy attempt. Surely there must be an easier way than needing to solve a recurrence relation. Any help appreciated.(I don't have the answer)

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lurflurf
Homework Helper
geometric series
if |x|<1
$$\sum_{k=n}^\infty x^k=\frac{x^n}{1-x}$$

Thanks for your response lurflurf. The series I've got looks like an alternating series so can I just use that formula? I'm probably not seeing something very obvious but any further assistance would be good thanks.

Edit: Nevermind, I checked my book and it appears that I can apply that formula. Thanks again for the help.

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$$\sum_{k=n}^\infty x^k=\frac{1}{1-x}$$ for |x| < 1.