Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone help me find a Leibniz Series (alternating sum) that converges to $5$ ?

Does such a series even exist?

Thanks in advance!!!

I've tried looking at a series of the form $ \sum _ 1 ^\infty (-1)^{n} q^n $ which is a geometric series ... But I get $q>1 $ , which is impossible... Does someone have an idea?

share|cite|improve this question
Do you know the sum of some alternating series? – AD. Jan 28 '13 at 17:17
Try $1+q=1/5$... – AD. Jan 28 '13 at 17:18
$ then I get $q < 0 ... $ which gives me negative elements... – theMissingIngredient Jan 28 '13 at 17:22
up vote 1 down vote accepted

You can cook up an alternating series that hits any sum $S$ you like as follows.

  1. Pick a geometric series (with first term of 1) with geometric ratio $r$, with $-1<r<0$. Its sum will be ${1\over 1-r}$.

  2. This is also an alternating series since $\sum_{n=0}^\infty r^n=\sum_{n=0}^\infty(-1)^{n}|r|^n$.

  3. If you want the final sum to be $S$, then you just need to multiply the original geometric series by a constant $k$ so that $k\sum_{n=0}^\infty r^n=S$, but this implies $k=S(1+|r|)$.

  4. Thus, you use $$S(1+|r|)\sum_{n=0}^\infty r^n=S(1+|r|)\sum_{n=0}^\infty (-1)^{n}|r|^n, \quad -1<r<0.$$ It will be alternating and have sum $S$.

share|cite|improve this answer
Thanks a lot!!!!! ! – theMissingIngredient Jan 28 '13 at 17:42

Hint: $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\cdots=\frac{2}{3}$.

share|cite|improve this answer

Take any Leibniz sequence $x_n$, the series of which converges not to zero, say to $c$, and then consider the sequence $(\frac5c\cdot x_n)_n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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