Well! I was going through harmonic series from mathworld.worlfram I found harmonic numbers are really tough to calculate I was scribbling and wrote this thing $$\sum_{k = 1}^{x}H_k= \sum_{k =1}^{x}\left(\sum_{u = 1}^{k}\frac 1{u}\right)$$ How can I simplify these highly complicated things Howe can we find this thing At the first, we can't find $H_k$ itself and series over series making it a little hard I wonder how can we simplify this thing? It's just my curiosity....
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Well! then I liked this curiosity
I guess by solving you mean just to find only one harmonic number(say: $H_p)$ and finding the above $\sum\sum\frac 1{k}$
This is possible and I solved that by finding a pattern that only the last terms of each harmonic number are making previously calculated harmonic numbers different with only $\frac 1{k-1}$
But for the sake proof of my simplified result, I can provide an easy way
$$\begin{align*} S &=\sum_{k =1}^{x}H_k\\ &=\sum_{k =1}^{x}\sum_{n =1}^{k}\frac 1{n}\\ &=\sum_{k =1}^{x}\frac 1{n}\sum_{k =n}^{x}1\\ & = \sum_{k =1}^{x}\frac 1{n}(x-n+1)\\ & =\sum_{k =1}^{x}\frac {(x+1)-n}{n}\\ & =(x+1)\sum_{k =1}^{x}\frac 1{n} - \sum_{k =1}^{x}1\\ & = (x+1)H_x - x \end{align*}$$
Here, SH() & test() is our result expected which are equal.
Darshan P.
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Wonderful! Elegant! – Nov 14 '21 at 10:43
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1You can simplify code: sympy.harmonic() – Nov 14 '21 at 11:06
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@HarryJane Thanks! def test(x): ... ans1 = sum(list(sympy.harmonic(k) for k in range(1, x+1))) ... ans2 = (x+1)sympy.harmonic(x) - x ... print(bool(ans1==ans2)) – Darshan P. Nov 14 '21 at 11:08
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Would like to know how often do you use sympy module? – Nov 14 '21 at 11:10
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@HarryJane Hardly, It just for *prime numbers, gamma, &zeta function: sympy.isprime()* – Darshan P. Nov 14 '21 at 11:12
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Nice to meet you! From France – Nov 14 '21 at 11:13
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1@HarryJane You are texting an introvert!!!! – Darshan P. Nov 14 '21 at 11:14
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Let us continue this discussion in chat. – Nov 14 '21 at 11:17