Cheyanne Leigh

2021-03-12

Determine whether the series $\sum {a}_{n}$ an converges or diverges: Use the Alternating Series Test.
$\sum _{n=2}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{n}{\mathrm{ln}\left(n\right)}$

Obiajulu

Consider the series $\sum _{n=2}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{n}{\mathrm{ln}\left(n\right)}$
Take ${a}_{n}=\left(-1{\right)}^{n}\frac{n}{\mathrm{ln}\left(n\right)}$ and ${b}_{n}=\frac{n}{\mathrm{ln}\left(n\right)}$
The Alternating series test is stated below:
Suppose the series $\sum {a}_{n}$ and either ${a}_{n}=\left(-1{\right)}^{n}{b}_{n}$ or ${a}_{n}=\left(-1{\right)}^{n+1}{b}_{n}$ where ${b}_{n}\ge 0$ for all n. Then if,
1.$\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=0$
2. $\left\{{b}_{n}\right\}$ is a decreasing sequence
The series is convergent.
Check the first condition for series convergent.
$\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{n}{\mathrm{ln}\left(n\right)}\right)$
$=\frac{\mathrm{\infty }}{\mathrm{\infty }}$
The value of the limit is in the indeterminate form.
Apply LHopitals rule to find the limit as follows.
$\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=lim\left(\frac{\frac{d}{dn}\left(n\right)}{\frac{d}{dn}\left(\mathrm{ln}\left(n\right)\right)}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{\frac{1}{n}}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(n\right)$
$=\mathrm{\infty }$
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 $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}\ne 0$ then the series $\sum {a}_{n}$ will diverge".
Since $\underset{n\to \mathrm{\infty }}{lim}{b}_{n}\ne 0$, the limit of the function ${a}_{n}=\left(-1{\right)}^{n}\frac{n}{\mathrm{ln}\left(n\right)}$ also not equal to zero as x goes to zero. That is, $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}\ne 0$.
By divergence test, it is concluded that the alternating series $\sum _{n=2}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{n}{\mathrm{ln}\left(n\right)}$ diverges.

Jeffrey Jordon