Consider the series \(\sum_{n=2}^\infty(-1)^n\frac{n}{\ln(n)}\)

Take \(a_n=(-1)^n\frac{n}{\ln(n)}\) and \(b_n=\frac{n}{\ln(n)}\)

The Alternating series test is stated below:

Suppose the series \(\sum a_n\) and either \(a_n=(-1)^nb_n\) or \(a_n=(-1)^{n+1}b_n\) where \(b_n\geq0\) for all n. Then if,

1.\(\lim_{n\to\infty}b_n=0\)

2. \(\left\{b_n\right\}\) is a decreasing sequence

The series is convergent.

Check the first condition for series convergent.

\(\lim_{n\to\infty}b_n=\lim_{n\to\infty}(\frac{n}{\ln(n)})\)

\(=\frac{\infty}{\infty}\)

The value of the limit is in the indeterminate form.

Apply L'Hopital's rule to find the limit as follows.

\(\lim_{n\to\infty}b_n=\lim(\frac{\frac{d}{dn}(n)}{\frac{d}{dn}(\ln(n))})\)

\(=\lim_{n\to\infty}(\frac{1}{\frac{1}{n}})\)

\(=\lim_{n\to\infty}(n)\)

\(=\infty\)

Observe that, the limit of the sequence goes to infinity as x goes to infinity. Thus, the series does not converges.

The divergence test states that, "If \(\lim_{n\to\infty}a_n\ne0\) then the series \(\sum a_n\) will diverge".

Since \(\lim_{n\to\infty}b_n\ne0\), the limit of the function \(a_n=(-1)^n\frac{n}{\ln(n)}\) also not equal to zero as x goes to zero. That is, \(\lim_{n\to\infty}a_n\ne0\).

By divergence test, it is concluded that the alternating series \(\sum_{n=2}^\infty(-1)^n\frac{n}{\ln(n)}\) diverges.

Take \(a_n=(-1)^n\frac{n}{\ln(n)}\) and \(b_n=\frac{n}{\ln(n)}\)

The Alternating series test is stated below:

Suppose the series \(\sum a_n\) and either \(a_n=(-1)^nb_n\) or \(a_n=(-1)^{n+1}b_n\) where \(b_n\geq0\) for all n. Then if,

1.\(\lim_{n\to\infty}b_n=0\)

2. \(\left\{b_n\right\}\) is a decreasing sequence

The series is convergent.

Check the first condition for series convergent.

\(\lim_{n\to\infty}b_n=\lim_{n\to\infty}(\frac{n}{\ln(n)})\)

\(=\frac{\infty}{\infty}\)

The value of the limit is in the indeterminate form.

Apply L'Hopital's rule to find the limit as follows.

\(\lim_{n\to\infty}b_n=\lim(\frac{\frac{d}{dn}(n)}{\frac{d}{dn}(\ln(n))})\)

\(=\lim_{n\to\infty}(\frac{1}{\frac{1}{n}})\)

\(=\lim_{n\to\infty}(n)\)

\(=\infty\)

Observe that, the limit of the sequence goes to infinity as x goes to infinity. Thus, the series does not converges.

The divergence test states that, "If \(\lim_{n\to\infty}a_n\ne0\) then the series \(\sum a_n\) will diverge".

Since \(\lim_{n\to\infty}b_n\ne0\), the limit of the function \(a_n=(-1)^n\frac{n}{\ln(n)}\) also not equal to zero as x goes to zero. That is, \(\lim_{n\to\infty}a_n\ne0\).

By divergence test, it is concluded that the alternating series \(\sum_{n=2}^\infty(-1)^n\frac{n}{\ln(n)}\) diverges.