$1+\dfrac{1}{9}+\dfrac{1}{45}+\dfrac{1}{189}+\dfrac{1}{729}+\dots=\sum\limits_{n=1}^\infty \dfrac{1}{(2n-1)\cdot 3^{n-1}}$
I got:
$\sum\limits_{n=1}^\infty \dfrac{1}{(2n-1)\cdot 3^{n-1}}=\sum\limits_{n=1}^\infty \dfrac{\int\limits_0^1 x^{2n-2}\,dx}{3^{n-1}}=\dots$
And no idea how to impove.
Thx!
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