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Examine whether the series defined as:

$$\sum a_n = \begin{cases} \hphantom{-}\dfrac{1}{n}, & \text{for $n$ odd} \\[8pt] -\dfrac{1}{5^n}, & \text{for $n$ even } \end{cases}$$

converges.

It is some kind of alternating series and the limit test seems to give convergence, as both branches $\lim\limits_{x \to \infty} \rightarrow$ $0$ and $a_n > a_{n+1}$ , i.e. decreacing. Is all the above paragraph right?

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Note that when an alternating series converges conditionally, it is because the series consisting only of the positive terms AND the series consisting of the negative terms both diverge. If the series of positive terms went to $+\infty$ while the series of negative terms converged to a fixed negative number $L < 0$, then the result goes to $+\infty$. An analogous reasoning shows what happens when the series of positive terms converge and the series of negative terms goes to $-\infty$.

In symbols, if $\sum_{n=1}^{\infty} a_n$ satisfies $a_{2n} > 0$ and $a_{2n+1} < 0$, then $$ a_{2n} \to +\infty, \quad a_{2n+1} \to L \in \mathbb R \quad \Longrightarrow \quad \sum_{n=1}^N a_n \underset{N \to \infty}{\longrightarrow} +\infty. $$ This can be proven formally since $$ \sum_{n=1}^{2N} a_n , \sum_{n=1}^{2N+1} a_n\ge \sum_{n=1}^N a_{2n} + \sum_{n=1}^{\infty} a_{2n+1} = \left( \sum_{n=1}^N a_{2n} \right) - L \underset{N \to \infty}{\longrightarrow} +\infty. $$ In your case, since the series $\sum_{n=1}^{\infty} \frac 1{2n-1}$ diverges and $\sum_{n=1}^{\infty} \frac{-1}{5^{2n}}$ converges, we can use the above reasoning. It is interesting to not simply look at the example but think a bit more thoroughly and learn something from the example, which is the point I tried to make above.

(I assumed the series converges conditionally ; when it converges unconditionally, both the series of positive terms and negative terms are convergent, so this whole business is not necessary.)

Hope that helps,

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    $\begingroup$ Great answer! It covered exactly the area in doubt. $\endgroup$ – Ziezi Sep 16 '16 at 16:06
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Note $$\sum_{i=1}^{\infty}a_{n}=\sum_{i=1}^{\infty}\frac{1}{2n-1}-\sum_{i=1}^{\infty}\frac{1}{5^{2i}}$$

Note that $$\sum_{i=1}^{\infty}\frac{1}{2n-1}$$ diverges, while $$\sum_{i=1}^{\infty}\frac{1}{5^{2i}}$$ converges. Thus your sum must diverge.

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  • $\begingroup$ @qbert Which means what exactly? $\endgroup$ – S.C.B. Sep 16 '16 at 15:37
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Your sum can be written as two sums $$ \sum_{n=1}^{\infty}\frac{1}{2n-1}-\sum_{n=1}^{\infty}\frac{1}{5^{2n}} $$ One of which diverges by comparison the the harmonic series, and the other is finite as it is geometric with appropriately small ratio.

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  • $\begingroup$ If the second sum was added to the first is fairly easy, but this minus sign confuses me as I can't estimate possible term cancellations. $\endgroup$ – Ziezi Sep 16 '16 at 15:38
  • $\begingroup$ It should be $\frac{1}{2n-1}$, not$ \frac{1}{2n+1}$! $\endgroup$ – S.C.B. Sep 16 '16 at 15:39
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    $\begingroup$ Good call will edit. It doesn't matter, in the end it is $\infty-c$ where $c$ is some constant $\endgroup$ – qbert Sep 16 '16 at 15:40
  • $\begingroup$ @Ziezi Not a problem, good luck $\endgroup$ – qbert Sep 16 '16 at 15:41

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