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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?

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Do you know the sum of some alternating series? –  AD. Jan 28 '13 at 17:17
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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

3 Answers 3

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 $S$.

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Thanks a lot!!!!! ! –  theMissingIngredient Jan 28 '13 at 17:42

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

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

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