$\displaystyle \sum^{\infty}_{n=3} \frac{cos(n\pi)}{n^{\frac{3}{5}}}$ What would I compare this to? $\displaystyle \frac{cos(n)}{n}$ ? Also, when n=3 how does this change things? Thanks!
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Notice that: $\displaystyle \cos (n\pi) = (-1)^{n}$ So really, your series is just: $\displaystyle \sum_{n=3}^{\infty} \frac{(-1)^n}{n^{\frac{3}{5}}}$ And there really isn't anything special about starting from $\displaystyle n = 3$.
Originally Posted by o_O Notice that: $\displaystyle \cos (n\pi) = (-1)^{n}$ So really, your series is just: $\displaystyle \sum_{n=3}^{\infty} \frac{(-1)^n}{n^{\frac{3}{5}}}$ And there really isn't anything special about starting from $\displaystyle n = 3$. Ok, so I would say that: $\displaystyle \sum a_n > 0$ so it converges and since $\displaystyle 0 < \sum a_{n+1} < \sum a_n$, then $\displaystyle \sum a_{n+1}$ also converges?
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