# How to prove these

How to prove these

(1) $\displaystyle\sum_{n=4}^\infty\frac{(-1)^n\ln n}n$ if it is absolutely or conditionally convergent? and

(2) $\displaystyle\sum_{n=1}^\infty\frac n{(-2)^n}$ if it is absolutely or conditionally convergent?

What I am thinking about the first one is that the series will be conditionally convergent as the sequence $\frac{\ln n}n$ is eventually decreasing to 0 and by Leibniz theorem the series will be convergent, but I don't know how show it not abosultely. And also I am thinking in the same thing about the second series.

Can any one solve these for me so that I can get the complete idea and I'll try to solve another problems by myself.

Thanks.

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Note that $|\ln n/n|\geq 1/n$ for $n\geq 3$. And for the second one, the ratio test can help you prove absolute convergence. –  julien May 2 at 4:15
A general strategy for dealing with infinite series is to test for divergence first (is $\lim_{n \rightarrow \infty} a_n \ne 0$?) . If the series passes that test ($\lim_{n \rightarrow \infty} a_n = 0$) , test for absolute convergence first; if the series is absolutely convergent, it is automatically conditionally convergent, and you're done. If the series does not converge absolutely, use the Alternating Series Test: if it passes, it's conditionally convergent; otherwise, it diverges. –  RecklessReckoner May 2 at 4:21
but how can I say that |ln(n)/n| is >= 1/n? I mean how can I know this true eventually? –  LoveMath May 2 at 4:22
At what value of $n$ is $\ln n > 1$? –  RecklessReckoner May 2 at 4:23
For $n\ge 3$, $\ln n >1$. –  vadim123 May 2 at 4:25
$(1)$ Compare the first series with the harmonic series whose general term is $\dfrac 1n$. Note that $$\;\left|\dfrac {\ln n}{n}\right| \geq \dfrac 1n \;\;\forall n\geq 3,$$ and we know the harmonic series diverges. Hence the first series converges conditionally (by the alternating series test), but not absolutely.
$(2)$ For the second series, try using the root test to prove absolute convergence. One can immediately see that $$\lim_{n\to \infty}\sqrt[\large n]{\left|\dfrac {n}{(-2)^n}\right|} = 1/2.$$ And we know that absolute convergence $\implies$ convergence, period.
ok, but how can we know that $\;\left|\dfrac {\ln n}{n}\right| \geq \dfrac 1n$? –  LoveMath May 2 at 4:27
Because for all terms $n \geq 3$, $\ln n \geq 1$. Since $n\to \infty$, we have infinitely many terms such that $(\ln n) / n \geq 1/n$. Just graph both $\;\left|\dfrac {\ln n}{n}\right|$ and $\dfrac 1n$ to see for yourself! ;-) –  amWhy May 2 at 4:30