Find the sum $sum_{n=1}^{infty} frac{x^{n+1}}{(n+1)*(n+2)}$
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Need to find the sum $sum_{n=1}^{infty} frac{x^{n+1}}{(n+1)(n+2)}$ What I did based on the suggestions: Multiplied and divided by $x$ "forgot" the $x$ in the denominator and then took the derivative of the remaining expression 2 times: ended up getting $frac{1}{x}sum_{n=1}^{infty} x^n$ Expanded this geometric series and then integrated the result once: $frac{1}{x}sum_{n=1}^{infty} x^n = frac{1}{x}int frac{x}{1-x}dx = frac{1}{x}(-x - ln(1-x)) $ And then integrated again: $frac{1}{x}int (-x - ln(1-x))dx = frac{1}{x}(-0.5x^2+x-xln(1-x)+ln(1-x)) $ which is apparently a correct answer. Is there any other more adequate way of solving this? Seems really non-intuitive to me, especially the double derivative-double integral part. Or maybe someone could shed the light regardi...