Question

Math

Posted 4 months ago

```
We know that
$\frac{n^{2}}{n^{3}-\ln (n)} \geq \frac{n^{2}}{n^{3}}=\frac{1}{n}>0$
for any $n \geq 1$.
Considering this fact, what does the direct comparison test say about
$\sum_{n=1}^{\infty} \frac{n^{2}}{n^{3}-\ln (n)} ?$
Choose 1 answer:
(A) The series converges.
(B) The series diverges.
(c) The test is inconclusive.
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

Consider the comparison series $\sum_{n=1}^{\infty} \frac{1}{n}$, which is the harmonic series

step 2

The harmonic series is known to diverge

step 3

Given that $\frac{n^{2}}{n^{3}-\ln (n)} \geq \frac{n^{2}}{n^{3}} = \frac{1}{n}$ for $n \geq 1$, we can compare the given series to the harmonic series

step 4

By the direct comparison test, if a series with non-negative terms is greater than or equal to a divergent series, then the original series also diverges

step 5

Since $\sum_{n=1}^{\infty} \frac{n^{2}}{n^{3}-\ln (n)} \geq \sum_{n=1}^{\infty} \frac{1}{n}$ and the harmonic series diverges, the given series also diverges

Answer

(B) The series diverges.

Key Concept

Direct Comparison Test

Explanation

If a series with non-negative terms is greater than or equal to a divergent series, then the original series also diverges. In this case, the given series is compared to the harmonic series, which is known to diverge, thus the given series also diverges.

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