EDIT: I want to prove it with &quot;converges absolutely&quot; and not something else. I want to prove that the sequance $\sum _{n=1}^{\infty \ }\left[n^3\sin \left(\frac{1}{n}\right)-n^2+\frac{1}{6}\right]$ converges. I thought of trying converges absolutely, but I don't even know how to start. Also, trying to derivate the inside function doesn't help showing that the function is a non-negative one. Any ideas? Thanks!
Let $t\in \mathbb R$. Prove that $\displaystyle{\sum_{n=1}^{\infty}n^{3}\sin\left(\frac{1}{n}\right)-n^{2}+t}$ converges iff $t=\frac{1}{6}$.  While the $(\Rightarrow)$ direction was fairly easy (as $a_n$ should converge to $0$, forcing $t=\frac{1}{6}$), i'm facing some hard time trying to prove the second direction, that is $\displaystyle{\sum_{n=1}^{\infty}(n^{3}\sin\left(\frac{1}{n}\right)-n^{2}+\frac{1}{6})}$ converges. I did notice that $a_n=a_{-n}$ and tried to write this series as a Leibnitz series, but had no luck since the sequence doesn't decrease. I also tried to prove absolute convergence but then again, no luck. Thanks.
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