Convergence of the Series $(-1)^n ln(n) / n$

To determine the convergence of the series $sum_{n1}^{infty} frac{(-1)^n ln(n)}{n}$, we can apply the Alternating Series Test. Let's explore this step by step.

Alternating Series Test

The Alternating Series Test states that a series of the form $sum (-1)^n a_n$ converges if two conditions are met:

$a_n geq a_{n 1}$ for all $n$, the sequence is non-increasing. $lim_{n to infty} a_n 0$.

In our case, we let $a_n frac{ln(n)}{n}$.

Step 1: Check the Limit $lim_{n to infty} a_n$

We need to compute the limit:

$lim_{n to infty} a_n lim_{n to infty} frac{ln(n)}{n}$

Using L'H?pital's Rule, since both the numerator and denominator approach infinity:

$lim_{n to infty} frac{ln(n)}{n} lim_{n to infty} frac{frac{1}{n}}{1} lim_{n to infty} frac{1}{n} 0$

So $lim_{n to infty} a_n 0$.

Step 2: Check if $a_n$ is Non-Increasing

We need to check if $a_n$ is non-increasing, which means $a_n geq a_{n 1}$.

Consider $a_n frac{ln(n)}{n}$. To analyze this, we can rewrite it as:

$frac{ln(n)}{n} geq frac{ln(n 1)}{n 1}$

This simplifies to:

$ln(n)(n 1) geq ln(n 1)n$

This inequality can be challenging to prove directly. Instead, let's analyze the function $f(x) frac{ln(x)}{x}$. The derivative $f'(x)$ can be computed using the quotient rule:

$f'(x) frac{1 - ln(x)}{x^2}$

Setting $f'(x) 0$ gives $ln(x) 1$ or $x e$. For $x e$, $f'(x) 0$ indicating the function is decreasing, and for $x e$, $f'(x) 0$ indicating the function is increasing. Thus, $a_n$ is non-increasing for $n geq 3$.

Conclusion

Since $a_n$ satisfies both conditions of the Alternating Series Test:

$lim_{n to infty} a_n 0$ $a_n$ is eventually non-increasing

We conclude that the series $sum_{n1}^{infty} frac{(-1)^n ln(n)}{n}$ converges.

$sum_{n1}^{infty} frac{(-1)^n ln(n)}{n} frac{ln^2 2}{2} - Gamma ln 2$

Using the Leibniz Test, we find that $ln(n)/n to 0$ as $n to infty$. If we consider the function $f(x) x^{1/x}$, it attains its maximum at $e$ and $x$ is monotonically decreasing if $x e$. Therefore, $ln(n)/n ln(n^{1/n})$ is monotonically decreasing if $n geq 4$. Thus, we can apply the Leibniz Test and conclude that the given series is convergent.