So last night i was sitting about bored and got to thinking about summations

Question

hba · Answer

And....

amistre64 · Answer

ugh, missed a "k" lol

$$\Large\sum_{k=1}^{n}\frac{1}{k}=\frac{\sum_{i=1}^{n}\frac{\prod_{k=1}^{n}k}{i}}{\prod_{i=k}^{n}k}$$

amistre64 · Answer

this got me thinking on how to do:$$\sum\frac{1}{an+k}$$for a hypergeometric series

amistre64 · Answer

$$\large \prod_{i=1}^{n}ai+k=...$$
i dont have a good notations for it, but it follows a pattern related to the binomial thrm

$$ \prod_{i=1}^{4}ai+k=4!a^4+(1.2.3+1.2.4+1.3.4+2.3.4)a^3k$$$$+(1.2+1.3+1.4+2.3+2.4+3.4)a^2k^2$$$$+(1+2+3+4)ak^3+k^4$$

amistre64 · Answer

the coeffs pattern out to all the combos tha the n! uses  ... if that makes any sense

amistre64 · Answer

$$y = \prod_{n=1}^{5}4n+3 $$
$$ln(y) = \sum_{n=1}^{5}ln(4n+3)$$

anonymous · Answer

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