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Given a formula for $\pi$ like:

$$4⋅\sum^\infty_{k=1} \frac{(−1)^{k+1}}{2k−1} = 4⋅(1−1/3+1/5−1/7+1/9−1/11…).$$

or some of the several others; how can you know that it holds true to any $k$? Couldn't it be that from a certain value upwards, the series start to diverge from $\pi$?

Robert Israel
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Pierre B
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  • What does 'any $k$' mean in this context? – Yuriy S May 31 '16 at 22:02
  • The series is convergent by Leibniz' criterion, hence it cannot diverge. – Jack D'Aurizio May 31 '16 at 22:02
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    The OP means "How we do we know that the series converges to $\pi$ (and not just something very close to $\pi$)?" The answer is by proving it :P – Eff May 31 '16 at 22:06
  • What does "how can you know that it holds true to any $k$?" mean? The series on the LHS is a sum of infinitely many terms (in a specific order), adding one at a time. The more you add, the closer you get to the limit value, which is $\pi$. – Clement C. May 31 '16 at 22:06
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    I think what you need to understand is this: we always prove such formulas by showing that they are equivalent to a certain definition for $\pi$ (or any other constant). We are never satisfied with adding a finite number of terms and thinking 'well, this looks like $\pi$' :) – Yuriy S May 31 '16 at 22:07
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    For example, if we say that $\pi$ is precisely the area of a cirlce with radius $1$, then we can define $\pi$ by the following integral: $$\pi=2 \int_{-1}^1 \sqrt{1-x^2}dx$$ Then any alleged 'formula for $\pi$' will need to be checked against this definition. If we can somehow show that one follows from another, then we are set – Yuriy S May 31 '16 at 22:12

1 Answers1

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The answer depends on where you want to start.

You can start with the definition of $\arctan x$, develop the series expansion for that, and then plug in $x=1$ to set the whole thing equal to $\pi/4$. Or you can calculate it geometrically the way Leibniz did.

Absent equating the series to some multiple of $\pi$ you can calculate two series that approach $\pi$ from above and below, and then squeeze out an expression for $\pi$ that way.

But the answer to your question is that often the series pops out from some expression which is already known to be a multiple of $\pi$.

John
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