Determine if $\sum_{t=1}^\infty (-1)^{n+1}\frac{(-4)^n}{n4^n}$ converges or diverges.

Question

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?

Answer 1

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.