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.

Question

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.

Accepted Answer

∻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.⚹◊What is the direct comparison test and how is it applied to infinite series involving fractions with n terms?⍭ Generate me a similar question◊

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