Question

question 10 of 13 determine convergence or divergence of $sum_{n = 17}^{infty}\frac{sqrt3{n}}{11n + 23}$ using any method covered so far. the series diverges. the series converges.

Explanation:

Step1: Use the limit - comparison test

Let \(a_n=\frac{\sqrt[3]{n}}{11n + 23}\) and choose \(b_n=\frac{n^{\frac{1}{3}}}{11n}=\frac{1}{11n^{\frac{2}{3}}}\).

Step2: Calculate the limit \(\lim_{n

ightarrow\infty}\frac{a_n}{b_n}\)

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Step3: Analyze the series of \(b_n\)

The series \(\sum_{n = 1}^{\infty}b_n=\sum_{n = 1}^{\infty}\frac{1}{11n^{\frac{2}{3}}}\) is a \(p\) - series with \(p=\frac{2}{3}<1\). A \(p\) - series \(\sum_{n = 1}^{\infty}\frac{1}{n^p}\) diverges when \(p\leqslant1\).

Step4: Draw the conclusion

Since \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=1>0\) and \(\sum_{n = 1}^{\infty}b_n\) diverges, by the limit - comparison test, \(\sum_{n = 17}^{\infty}a_n=\sum_{n = 17}^{\infty}\frac{\sqrt[3]{n}}{11n + 23}\) diverges.

Answer:

The series diverges.