$\displaystyle \sum {\frac {2+sin^3(n+1)} {2^n + n^2} }$ Can someone show me this plz?
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Originally Posted by Cato $\displaystyle \sum {\frac {2+sin^3(n+1)} {2^n + n^2} }$ Can someone show me this plz? note that $\displaystyle \Bigg| \frac {2 + \sin^3 (n + 1)}{2^n + n^2} \Bigg| \le \Bigg| \frac {2 + 1}{2^n + n^2} \Bigg| = \Bigg| \frac 3{2^n + n^2} \Bigg|$ now what can you say?
$\displaystyle \Bigg| \frac 3{2^n + n^2} \Bigg|$ < $\displaystyle \Bigg| \frac 3{n^2} \Bigg|$ which we know converges so the whole thing converges, right?
Last edited by Cato; Oct 9th 2008 at 11:17 PM.
Originally Posted by Cato $\displaystyle \Bigg| \frac 3{2^n + n^2} \Bigg|$ < $\displaystyle \Bigg| \frac 3{n^2} \Bigg|$ which we know converges so the whole thing converges, right? yes, it converges absolutely by the comparison test. and absolute convergence implies convergence
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