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I've been doing a lot of stuff with infinite series and Python lately, and have been getting some very interesting results. For example, the series 1 + 1/16 - 1/256 - 1/4096 + ... (two plus signs, then two minus signs, etc.) converges to 272/257 - wow! But the series 1 + 1/16 + 1/256 + 1/4096 - ... (four alternating plus and minus signs) converges to 69904/65537 - a very interesting result indeed.

I know the series 1 - 1/2 + 1/3 - 1/4 + ... is equal to ln 2. But the series 1 + 1/2 - 1/3 - 1/4 + 1/5 + ... is equal to pi/4 + (1/2 ln 2) - what a strange result!

I also tried a series involving triangular numbers. The series 1 + 1/3 + 1/6 + 1/10 + ... does not seem to converge. What about the series 1 - 1/3 + 1/6 - 1/10 + ... ? There HAS to be some sort of special result to that one, and after 10,000,000 iterations the result stabilized around 0.77258872. Can someone please help me figure this out? (I don't have any experience with calculus)

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marked as duplicate by Nosrati, Namaste calculus Aug 7 '18 at 0:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The alternating series theorem says that any sum of terms that alternate in sign and have magnitudes monotonically decreasing to zero converges. If your series $1+1/3+1/6+1/10+\ldots$is $\sum_{i=1}^\infty \frac 2{i(i+1)}$ it converges to $2$ $\endgroup$ – Ross Millikan Aug 6 '18 at 18:16
  • $\begingroup$ This has already been answered in your previous question. $0.77\ldots = 4\log 2-2$. $\endgroup$ – Jack D'Aurizio Aug 6 '18 at 18:17
  • $\begingroup$ Ok sorry i didnt see it in my other post, someone told me to create a new post $\endgroup$ – Anonymous Aug 6 '18 at 18:30
  • $\begingroup$ Ross Millikan I see now, there was an error in my original program. Thanks! $\endgroup$ – Anonymous Aug 6 '18 at 18:32
  • $\begingroup$ @Anonymous Delete this post! $\endgroup$ – Nosrati Aug 6 '18 at 18:47
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This is $$2\left(\frac11-\frac12-\frac12+\frac13+\frac13-\frac14-\cdots\right)=-2+4\ln2.$$

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Triangular numbers can be given by the formula: $$T_n=\frac{n(n+1)}{2}$$ Thus, $$\sum_{n=1}^\infty (T_n)^{-1}=2\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)=2.$$

Similarly, the sum of alternating triangular numbers can be determined (here you can look at groups of $4$ to handle the alternating sign as well as the partial fraction decomposition).

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