Proving that \(\displaystyle\sum^{\infty}_{\left\lbrace{n}={1}\right\rbrace}{\frac{{{\left(-{1}\right)}^{{{n}+{1}}}}}{{{n}}}}={\log{{2}}}\)

Jaylin Clements

Jaylin Clements

Answered question

2022-03-28

Proving that {n=1}(1)n+1n=log2

Answer & Explanation

arrebusnaavbz

arrebusnaavbz

Beginner2022-03-29Added 18 answers

Fom the Taylor expansion of the natural logarithm
ln(1+x)=n=1(1)n+1nxn
With x=1,{n=1}(1)n+1n=ln(2)
Ben Castillo

Ben Castillo

Beginner2022-03-30Added 13 answers

hint
Let f(x)=ln(1+x). f is C near 0. thus for n>0,c(0,1):
f(1)=f(0)+k=1n1k!f(k)(0)+1(n+1)!f(n+1)(c)
but f(n+1)(c)=(1)nn!(1+c)n+1
and 0<11+c<1 , so
limn+k=1n(1)k+1k=f(1)=ln(2).

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