Key to "Check the convergence or divergence of the series∑n=2∞(−1)n(1log⁡n+cos⁡nn2)"

Cormac Gilbert

Answered question

2022-02-28

Check the convergence or divergence of the series

\sum_{n = 2}^{\infty} {(- 1)}^{n} (\frac{1}{\log n} + \frac{\cos n}{n^{2}})

alagmamGynccip

Beginner2022-03-01Added 4 answers

\sum_{n = 2}^{\infty} {(- 1)}^{n} (\frac{1}{\ln (n) + \frac{\cos (n)}{n^{2}}}) = \sum_{n = 2}^{\infty} (\frac{{(- 1)}^{n}}{\ln (n)} + {(- 1)}^{n} \frac{\cos (n)}{n^{2}})

The series

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

is convergent by the alternating series test;

\frac{1}{\ln (n)}

is strictly decreasing and tends to 0 as

n \to \infty

. The series

\sum_{n = 2}^{\infty} {(- 1)}^{n} \frac{\cos (n)}{n^{2}}

is absolutely convergent because

| {(- 1)}^{n} \frac{\cos (n)}{n^{2}} | = \frac{| {(- 1)}^{n} | | \cos (n) |}{n^{2}} \leq \frac{1}{n^{2}}

and

\sum_{n = 2}^{\infty} \frac{1}{n^{2}}

is a convergent p-series. It follows that the sum of these two series

\sum_{n = 2}^{\infty} (\frac{{(- 1)}^{n}}{\ln (n)} + {(- 1)}^{n} \frac{\cos (n)}{n^{2}}) = \sum_{n = 2}^{\infty} {(- 1)}^{n} (\frac{1}{\ln (n)} + \frac{\cos (n)}{n^{2}})

is convergent.

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