The n-th harmonic number is defined by
Hn=∑n, i=1 (1/i) =1+1/2+1/3+…+1/n.
Show that
Hn=∫(1>0) 1−x^n/1−x dx.
"x=1-u"
"\\int^1_0\\frac{1-x^n}{1-x}dx=-\\int^0_1\\frac{1-(1-u)^n}{u}du=\\int^1_0\\frac{1-(1-u)^n}{u}du="
"=\\int^1_0[\\displaystyle{\\sum_{k=1}^n(-1)^{k-1}\\begin{pmatrix}\n n \\\\\n k \n\\end{pmatrix}u^{k-1}}]du=\\displaystyle{\\sum_{k=1}^n(-1)^{k-1}}\\begin{pmatrix}\n n \\\\\n k \n\\end{pmatrix}\\int^1_0u^{k-1}du="
"=\\displaystyle{\\sum_{k=1}^n(-1)^{k-1}}\\frac{1}{k}\\begin{pmatrix}\n n \\\\\n k \n\\end{pmatrix}"
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