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sum_{n=1}^{infty} [sqrt(n^4+1)/(n^3+n^2)] with steps
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\[\sum_{n=1}^{\infty} \sqrt(n^4+1)/(n^3+n^2)\]
\[\sum_{n=1}^{\infty} \sqrt{\frac{n^4+1}{n^3+n^2}}\]
no the square root sign is only for the numerator
oh ok..
\[\sum_{n=1}^{\infty}\frac {\sqrt{n^4+1}}{n^3+n^2}\]
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yes that's it
http://www.wolframalpha.com/input/?i=sum+sqrt%28n^4%2B1%29%2F%28n^3%2Bn^2%29%2C+n%3D1+to+infinity
thanks, i know how do do that but i need the steps of how to do it
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