I found pic related written in the back of an old reference book,

>>8613331
I don't know what the fuck is going on because in the first term where n=1, you get a division by 0.

>>8613345
my mistake, the second factor in the denom
should be (n + 1)

File: pi-nilakantha.png (5KB, 456x60px) Image search: [Google]

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this is the similar Nilakantha expression
[math] \displaystyle \pi =3-4 \sum_{n \geq 1} \frac{(-1)^n}{2n(2n+1)(2n+2)}[/math]

File: answer.gif (2KB, 250x48px) Image search: [Google]

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>>8613331

The summation would have to converge to 6.14159 or -.14159...

.14159 it is!

second try at the Nilakantha expression:
[math] \displaystyle \pi = 3-4 \sum_{n \geq 1} \frac{(-1)^n}{2n(2n+1)(2n+2)}[/math]

>>8613375
Then why do you need us to confirm? You can prove it using that.

the nth term of that sequence is
[eqn] - \frac{4 (-1)^n}{2n(2n+1)(2n+2)} = - \frac{2 (-1)^n}{n(2n+1)(2n+2)} = \frac{2 (-1)^n}{n(2n+1)2(n+1)} = \frac{(-1)^n}{n(2n+1)(n+1)} [/eqn]

Basic algebra, my man.

>>8613385
>>8613375

Oops, the minus sign should be in all the equalities, let me fix that.

[eqn] - \frac{4 (-1)^n}{2n(2n+1)(2n+2)} = - \frac{2 (-1)^n}{n(2n+1)(2n+2)} = - \frac{2 (-1)^n}{n(2n+1)2(n+1)} = - \frac{(-1)^n}{n(2n+1)(n+1)} [/eqn]

← Nilakantha expression
[math] \displaystyle \pi =3+ \frac{4}{2 \times 3 \times 4}- \frac{4}{4 \times 5 \times 6}+ \frac{4}{6 \times 7 \times 8}- \cdots[/math]

>>8613389
Thank you for showing this, I've been having
a Hell of a time tonight and just couldn't see it.

File: pi-series.gif (1KB, 310x68px) Image search: [Google]

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I combined pairs of terms to get a faster-converging expression:
[math] \displaystyle \pi =3+ \frac{6}{3 \times 15}+ \frac{6}{15 \times 63}+ \frac{6}{35 \times 143}+ \cdots[/math]

>>8614697
all terms positive so the series approaches
the limit from below, giving an opportunity to
estimate the truncation error