Determine whether the series sum a_n an converges or diverges: Use the Alternating Series Test. sum_{n=2}^infty(-1)^nfrac{n}{ln(n)}

Cheyanne Leigh 2021-03-12 Answered
Determine whether the series \(\sum a_n\) an converges or diverges: Use the Alternating Series Test.
\(\sum_{n=2}^\infty(-1)^n\frac{n}{\ln(n)}\)

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Expert Answer

Obiajulu
Answered 2021-03-13 Author has 15215 answers
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.
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