Questions: Use the Comparison Test to determine whether the infinite series is convergent. sum from n=2 to infinity of n^(9/8) / (n^2 - n) By the Comparison Test, the infinite series sum from n=2 to infinity of n^(9/8) / (n^2 - n) A. converges B. diverges

Solution Steps

To determine whether the given infinite series converges or diverges using the Comparison Test, we need to compare it to a known series. We can simplify the given series and compare it to a p-series.

1. Simplify the given series: \(\frac{n^{\frac{9}{8}}}{n^2 - n}\).
2. For large \(n\), \(n^2 - n \approx n^2\), so the series behaves like \(\frac{n^{\frac{9}{8}}}{n^2} = n^{\frac{9}{8} - 2} = n^{-\frac{7}{8}}\).
3. Compare this to the p-series \(\sum \frac{1}{n^p}\) where \(p = \frac{7}{8}\).
4. Since \(p = \frac{7}{8} < 1\), the p-series diverges.
5. By the Comparison Test, if a series is greater than or comparable to a divergent series, it also diverges.

We start with the series

\[ \sum_{n=2}^{\infty} \frac{n^{\frac{9}{8}}}{n^{2}-n}. \]

For large \( n \), we can approximate the denominator:

\[ n^2 - n \approx n^2. \]

Thus, the series behaves like

\[ \frac{n^{\frac{9}{8}}}{n^2} = n^{\frac{9}{8} - 2} = n^{-\frac{7}{8}}. \]

We compare our series to the p-series

\[ \sum_{n=2}^{\infty} \frac{1}{n^{\frac{7}{8}}}. \]

Since \( p = \frac{7}{8} < 1 \), the p-series diverges.

By the Comparison Test, if

\[ \frac{n^{\frac{9}{8}}}{n^{2}-n} \geq \frac{1}{n^{\frac{7}{8}}} \]

for sufficiently large \( n \), then our series also diverges.

For \( n = 1000 \):

• The term value is

\[ \frac{1000^{\frac{9}{8}}}{1000^{2}-1000} \approx 0.00237374745311477. \]

• The corresponding p-series term is

\[ \frac{1}{1000^{\frac{7}{8}}} \approx 0.0023713737056616554. \]

Since

\[ 0.00237374745311477 \geq 0.0023713737056616554, \]

the series diverges.

Final Answer