4
$\begingroup$

I have this series:

$$S_n = \sum_{k=1}^\infty \frac{5}{k} = 5+\frac{5}{2} + \frac{5}{3} + .... + \frac{5}{k}$$

I wonder if theres a method to find the $r$ for any geometric series, Because sometimes I can't "see" it.

I thought of something like that :

I need to find $r$ from $5$ to $\frac{5}{2}$ hence,

$$5r=\frac{5}{2} \rightarrow r = \frac{1}{2}$$

The problem is that it works only for $S_2$.

However If I calculate for $S_3$ using this method I get wrong answer.

$$S_3=5+\frac{1}{2}*5+(\frac{1}{2})^2*5 = 5+\frac{5}{2}+\frac{5}{4} \neq 5+\frac{5}{2}+\frac{5}{3} = S_3$$

I wonder if there is a method that could work on every geometric series.

Any help will be appreciated.

$\endgroup$
  • 2
    $\begingroup$ This fails because the sequence is not geometric. The very definition of geometric series gives immediately that (for nonzero series) the ratio of two successive elements is the same. $\endgroup$ – Travis Willse Oct 28 '14 at 13:23
  • 2
    $\begingroup$ Harmonic Progressions don't have common ratio $\endgroup$ – user171358 Oct 28 '14 at 13:23
  • 2
    $\begingroup$ The above series is not geometric. $\endgroup$ – Jasser Oct 28 '14 at 13:24
  • 2
    $\begingroup$ You were right in your method to find $r$, by the way. But as others have pointed out, this isn't a geometric series. And you actually proved it because the $r$ you found from the first to second term isn't the same $r$ from the second to third term. $\endgroup$ – layman Oct 28 '14 at 13:25
  • 2
    $\begingroup$ This is a perfectly good question: the OP knows what he is asking, he has tried to solve the problem on his own and has posted his work. There is absolutely no reason to downvote this question! $\endgroup$ – 5xum Oct 28 '14 at 13:34
4
$\begingroup$

This is not Geometric Series

Your method of finding ratio of subsequent terms works perfectly for Every Geometric series

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.