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∑ (x^(k+3))/(k+1),k=0..∞ for the above taylor seriesabout the origin ,find the function . show steps plz.
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\[\sum_{k=2}^{\infty}(x ^{k+3})/(k+1)\]
\[k=0...\infty\quad\mbox{or}\quad k=2...\infty\quad ???\]
k=2..∞
\[\sum\limits_{k=2}^{\infty}\frac{x^{k+3}}{k+1}=x^2\sum\limits_{k=2}^{\infty}\frac{x^{k+1}}{k+1}=x^2\cdot S(x)\]\[S(x)=\sum\limits_{k=2}^{\infty}\frac{x^{k+1}}{k+1}\quad\Rightarrow\quad S'(x)=\sum\limits_{k=2}^{\infty}x^k=x^2+x^3+\cdots\]\[S'(x)=x^2\frac{1}{1-x}=\frac{x^2-1+1}{1-x}=-1-x+\frac{1}{1-x};-1\leq x<1\]\[S(x)=\int S'(x)dx=-x-\frac{x^2}{2}-\ln{(1-x)}\]\[\sum\limits_{k=2}^{\infty}\frac{x^{k+3}}{k+1}=-x^2\left(x+\frac{x^2}{2}+\ln{(1-x)}\right)\]
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