Determine whether the series converges or diverges.
(n=1→[infinity]) Σ (n + 1)/ (n⁹ + n)

a) The series converges by the Comparison Test. Each term is less than that of a convergent geometric series.
b) The series converges by the Comparison Test. Each term is less than that of a convergent p-series.
c) The series diverges by the Comparison Test. Each term is greater than that of a divergent p-series.
d) The series diverges by the Comparison Test. Each term is greater than that of a divergent harmonic series.

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LammettHash LammettHash · Answer 1 · 2020-05-01T18:44:15+02:00

[tex]\displaystyle\sum_{n=1}^\infty\frac{n+1}{n^9+n}[/tex]

B: Since

[tex]\dfrac{n+1}{n^9+n}=\dfrac{n+1}{n(n^8+1)}\approx\dfrac1{n^8}[/tex]

we can compare the given series to the convergent p-series,

[tex]\displaystyle\sum_{n=1}^\infty\frac1{n^7}[/tex]

We have

[tex]n^8<n^9[/tex]

[tex]\implies n^8+n^7<n^9+n[/tex]

(for large enough [tex]n[/tex], the degree-9 polynomial dominates the degree-8 polynomial)

[tex]\implies n^7(n+1)<n^9+n[/tex]

[tex]\implies\dfrac{n+1}{n^9+n}<\dfrac1{n^7}[/tex]