Find a closed form (not a sum) of the power series
$$f(x) = \sum^\infty_{n=0 } \frac{x^n}{n+1}$$
which converges on the interval $[-1,1)$ by differentiating the series on its interval of convergence and finding an antiderivative to lead to $f(x)$
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4The hint in the question could give you an answer. What happened when you tried it? Perhaps looking at $xf(x)$ might make it even easier – Henry Dec 12 '22 at 20:48
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what have you done? would you like to share your thoughts? – Math-fun Dec 12 '22 at 21:47
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Does this answer your question? How to find generating function of a sequence – Sine of the Time Dec 12 '22 at 23:40
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Notice that: $$xf(x)=\sum_{n\ge0}\int x^ndx$$Since this converges, flip the order of the integral and the sigma:$$\sum_{n\ge0}\int x^ndx=\int\sum_{n\ge0} x^ndx$$Use geometric series to get: $$\int\frac{1}{1-x}dx=xf(x)$$$$-\frac{1}{x}\ln(1-x)=f(x)$$Thanks to @Zakhurf for correcting my answer.
Kamal Saleh
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1The last part of this answer is wrong. We obtain $$x f(x) = \int \frac{1}{1-x}dx$$ which can be written as an elemantary function. Indeed $$f(x) = -\frac{1}{x}\ln(1-x)$$ However, the rest of the reasoning is correct. – Zakhurf Dec 12 '22 at 22:51
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