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I tried to use ratio and root test to see the convergence of $$\sum_{n=1}^{+\infty}\frac{(-1)^{n}3n}{4n-1}$$ but both were inconclusive.

I also tried to use Leibniz test. I got that $|a_{n+1}|\leq|a_{n}|$, but I don't know how to calculate this limit: $$\lim_{n\to\infty}a_{n}$$ I know the series does not converge, but I can't show it.

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    $\begingroup$ Hint: If $\sum b_n$ converges, then $\lim_{n\to\infty}b_n=0$. $\endgroup$ – André Nicolas Nov 29 '15 at 16:27
  • $\begingroup$ But how can I calculate this limit? I couldn't. $\endgroup$ – mvfs314 Nov 29 '15 at 16:28
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    $\begingroup$ The limit does not exist. The limit of the absolute values is $3/4$. For $n$ laege, $b_n$ bounces back and forth between close to $3/4$ and close to $-3/4$. $\endgroup$ – André Nicolas Nov 29 '15 at 16:30
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Note that $$\lim_{n\to\infty}\frac {(-1)^n3n}{4n-1}$$

does not exist, since it can be rearranged to

$$\lim_{n\to\infty}(-1)^n\frac {3n}{4n-1}$$

We can then easily see that as $x$ approaches $\infty$, it oscillates between positive and negative values, while $\frac{3n}{4n-1}$ approaches $\frac{3}{4}$. Since it does not converge to 0, we can conclude that the series also diverges.

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Hint: This series diverges since general term does not tend to zero.

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