Question

Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series. sum_{n=2}^inftyfrac{(-1)^nn}{n^2-3}

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asked 2020-10-25
Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series.
\(\sum_{n=2}^\infty\frac{(-1)^nn}{n^2-3}\)

Expert Answers (1)

2020-10-26

We have given series
\(\sum_{n=2}^\infty\frac{(-1)^nn}{n^2-3}\)
Let, \(\sum_{n=2}^\infty(-1)^na_n=\sum_{n=2}^\infty\frac{(-1)^nn}{n^2-3}\)
Then, \(a_n=\frac{n}{n^2-3}\)
\(a_{n+1}=\frac{n+1}{(n+1)^2-3}\)
\(\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}\frac{n}{n^2-3}\)
\(=\lim_{n\rightarrow\infty}\frac{n}{n^2(1-\frac{3}{n^2})}\)
\(=0\)
\(n^2-3<(n+1)^2-3\)
\(\frac{1}{n^2-3}>\frac{1}{(n+1)^2-3}\)
\(\frac{n}{n^2-3}>\frac{n}{(n+1)^2-3}\)
\(a_n>a_{n+1}\)
Therefore, sequence \(\left\{a_n\right\}\) is decreaing.
So, by alternating series test given series is convergent.

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