Series Convergence/Divergence (II)

[shamieh]

Determine whether the following series converges or diverges and justify your answer.

$$\sum^{\infty}_{n = 2} \frac{(-1)^n ln(n)}{n}$$

So can't I just apply L'opitals to obtain $$\frac{1}{n}$$ which diverges because it is a alternating harmonic series, thus: converges?

 

[Prove It]

Determine whether the following series converges or diverges and justify your answer.

$$\sum^{\infty}_{n = 2} \frac{(-1)^n ln(n)}{n}$$

So can't I just apply L'opitals to obtain $$\frac{1}{n}$$ which diverges because it is a alternating harmonic series, thus: diverges?

Not even close I'm afraid. L'Hospital's Rule is used to find the limit of a function, not anything to do with determining the convergence or divergence of a series.

This is an alternating series, so by LAST (Lagrange Alternating Series Test) the series will converge if the non-alternating part is decreasing.

Consider the function $\displaystyle \begin{align*} f(x) = \frac{\ln{(x)}}{x} \end{align*}$. We have $\displaystyle \begin{align*} f'(x) = \frac{x\left( \frac{1}{x} \right) - 1 \left[ \ln{(x)} \right] }{x^2} = \frac{1 - \ln{(x)}}{x^2} \end{align*}$. This will be negative for all $\displaystyle \begin{align*} x > e \end{align*}$, thereby showing that the function $\displaystyle \begin{align*} f(x) \end{align*}$ is decreasing for all $\displaystyle \begin{align*} x > e \end{align*}$.

This at least shows that $\displaystyle \begin{align*} \sum_{n = 3}^{\infty}{ \frac{(-1)^n \ln{(n)}}{n} } \end{align*}$ is convergent. If you add the extra term where $\displaystyle \begin{align*} n = 2 \end{align*}$, you will simply be adding two finite numbers, thereby still giving you something finite. So the series you have given is also convergent.

 

[shamieh]

So I can't do this ? :

$$\sum^{\infty}_{n = 2} (-1)^n \frac{ln(n)}{n} = n\to\infty \frac{1/n}{1} = 0$$

now since $$\lim a_n = 0 $$ using AST , $$0 < \frac{1}{n+1} < 1/n$$ so the series converges using alternating series test?

 

[chisigma]

Determine whether the following series converges or diverges and justify your answer.

$$\sum^{\infty}_{n = 2} \frac{(-1)^n ln(n)}{n}$$

So can't I just apply L'opitals to obtain $$\frac{1}{n}$$ which diverges because it is a alternating harmonic series, thus: converges?

The series converges for the Leibnitz's criterion. Remembering the definition of the Dirichlet Eta Function...

$\displaystyle \eta(s) = - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{s}} = (1 - 2^{1-s})\ \zeta(s)\ (1)$... where $\zeta(*)$ is the Riemann zeta function. Deriving (1) we obtain...

$\displaystyle \eta^{\ '}(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n}\ \ln n}{n^{s}} = 2^{1-s}\ \ln 2\ \zeta(s) + (1 - 2^{1-s})\ \zeta^{\ '} (s)\ (2)$

... and with a little of patience you can demonstrate that is $\displaystyle \eta^{\ '} (1) = \gamma\ \ln 2 - \frac{\ln^{2} 2}{2}$, where $\gamma = .5772...$ is the Euler's constant... Kind regards $\chi$ $\sigma$