Determine if $$\sum_{t=1}^\infty (-1)^{n+1}\frac{(-4)^n}{n4^n}$$ converges or diverges. To make it simpler to deal with, I managed to simplify the sum to $$\sum_{t=1}^\infty (-1)^{2n+1}\frac{1}{n}$$ but I can't seem to find a way to show if it diverges or converges. With the Ratio and Root test, the limit comes to 1, so it is inconclusive. What other ways would work?

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    $\begingroup$ $2n+1$ is odd, so you can simply further your sum to $$\sum_{n=1}^\infty\dfrac{-1}{n}.$$ This is the classic harmonic series. Can you proceed further? $\endgroup$ – Workaholic Mar 11 '16 at 22:59
  • $\begingroup$ @RecklessReckoner is living up to his name. (This isn't an alternating series.) $\endgroup$ – Christopher Carl Heckman Mar 11 '16 at 23:20
  • $\begingroup$ @CarlHeckman Whoops! I'm seeing powers of -1 where there aren't any. I withdraw my earlier comment (wonder who upvoted it?). $\endgroup$ – colormegone Mar 11 '16 at 23:23
  • $\begingroup$ On the matter of the Ratio Test, it is usually inconclusive when the general term only involves powers of $ \ n \ $ (which is what the term in your series became after factor cancellation). $\endgroup$ – colormegone Mar 11 '16 at 23:25

If you notice that $(-4)^n=(-1)^n\cdot 4^n$, you can rewrite the general term as $$ (-1)^{n+1}\frac{(-1)^n\cdot 4^n}{n\cdot 4^n}= (-1)^{2n+1}\frac{1}{n}=-\frac{1}{n} $$ Since a series with general term $a_n$ converges if and only if the series with general term $-a_n$ converges (proof?), you are reduced to see whether $$ \sum_{n=1}^{\infty}\frac{1}{n} $$ converges. Does it?

No, it is a well-known divergent series. You could prove it by observing that $$ 1+\frac{1}{2}+\dots+\frac{1}{n}\ge\log n $$ using Riemann sums, if you don't know about the harmonic series.

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    $\begingroup$ Or (my favorite proof), that the sum is $1+\frac{1}{2}+ \left(\frac{1}{3}+\frac{1}{4}\right) + \left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7} + \frac{1}{8}\right) + \cdots > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots$. $\endgroup$ – rogerl Mar 12 '16 at 0:05
  • $\begingroup$ @rogerl I wanted to make a change, for once. ;-) $\endgroup$ – egreg Mar 12 '16 at 0:10

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