On 2018-03-24 14:43, françois mendzina essomba2 via math-fun wrote:
Hello...
I noticed that :
atan(x)=(205*x^7+2184*x^5+5760*x^3+4096*x)/(36*x^8+820*x^6+4368*x^4+7680*x^2+4096)+sum(S(n),n=1..infinity);
with
S(m)=2^(-m-2)*x*sum(1/((2*k-1)^2*2^(-2*m-4)*x^2+1)-1/(k^2*2^(-2*m-2)*x^2+1),k,1,2^(m+1));
we deduce from it:
1)
Pi= 2449/850 + 4*sum(S(n),n=1..infinity);
with
S(m)=2^(-m-2)*sum(1/((2*k-1)^2*2^(-2*m-4)+1)-1/(k^2*2^(-2*m-2)+1),k,1,2^(m+1)) ;
[. . .]
Dr. Essomba: While In[11]:= Pi==2449/850+4*Sum[S@n,{n,\[Infinity]}] In[13]:= S@m_-
2^(-m-2)Sum[1/((2*k-1)^2*2^(-2*m-4)+1)-1/(k^2*2^(-2*m-2)+1),{k,2^(m+1)}]
Out[14]= S[m_]->1/8 (2^-m+2 I (2 I \[Pi] Csch[2^(2+m) \[Pi]]+HarmonicNumber[(1-I) 2^(1+m)]-HarmonicNumber[(1+I) 2^(1+m)]-PolyGamma[0,1/2+(1-I) 2^(1+m)]+PolyGamma[0,1/2+(1+I) 2^(1+m)])) In[15]:= %11/.% Out[15]= \[Pi]==2449/850+4 \!\( \*UnderoverscriptBox[\(\[Sum]\), \(n\), \(\[Infinity]\)]\( \*FractionBox[\(1\), \(8\)]\ \(( \*SuperscriptBox[\(2\), \(-n\)] + 2\ I\ \((2\ I\ \[Pi]\ Csch[ \*SuperscriptBox[\(2\), \(2 + n\)]\ \[Pi]] + HarmonicNumber[\((1 - I)\)\ \*SuperscriptBox[\(2\), \(1 + n\)]] - HarmonicNumber[\((1 + I)\)\ \*SuperscriptBox[\(2\), \(1 + n\)]] - PolyGamma[0, \*FractionBox[\(1\), \(2\)] + \((1 - I)\)\ \*SuperscriptBox[\(2\), \(1 + n\)]] + PolyGamma[0, \*FractionBox[\(1\), \(2\)] + \((1 + I)\)\ \*SuperscriptBox[\(2\), \(1 + n\)]])\))\)\)\) In[16]:= N@% Out[16]= True is numerically plausible, I'm afraid that under US law, this makes you a PolyGamist. --rwg