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$$a_{m} = \begin{cases} \dfrac{-2}{m+1}, &\text{ if m is odd}\\[1em] \dfrac{16}{m^4}, &\text{ if m is even} \end{cases}$$

If m is even the series should converge by P series test. When m is odd $a_{m} \approx \frac{-1}{m}$ which is a alternating harmonic series and it converges. What we can say series $$\sum_{m=3}^{\infty}a_{m}$$ as a whole?

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  • $\begingroup$ Shouldn't you include a sigma? $\endgroup$ – S.C.B. Mar 12 '16 at 11:44
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Show that $\sum_{n=3}^{\infty} a_n=-\infty$. \begin{align} \sum_{k=3}^{2n} a_k &= -\frac{1}{2}+\frac{1}{16}-\frac{1}{3}+\frac{1}{81}-\cdots-\frac{1}{n}+\frac{1}{n^4}\\ &<-\frac{1}{2}+\frac{1}{4}-\frac{1}{3}+\frac{1}{6}-\cdots-\frac{1}{n}+\frac{1}{2n}\\ &<-\frac{1}{4}-\frac{1}{6}-\cdots-\frac{1}{2n} \end{align} and $-\frac{1}{4}-\frac{1}{6}-\cdots-\frac{1}{2n}$ goes to $-\infty$ as $n\to\infty$. Therefore given series diverges.

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$$\sum_{m=3}^{\infty} a_m = \sum_{m=2}^{\infty}a_{2m} + \sum_{m=1}^{\infty} a_{2m + 1}$$

But

$$\sum_{m=2}^{\infty} a_{2m} = \sum_{m=2}^{\infty} \frac1{m^4}$$

converges.

And,

$$\sum_{m=1}^{\infty} a_{2m+1} = -\sum_{m=1}^{\infty} \frac1{m+1}$$

diverges.

Therefore the series as a whole diverges.

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