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Test the series for convergence or divergence: n=1n51n6+1.\displaystyle\sum_{n=1}^\infty \frac{n^5 - 1}{n^6 + 1}.

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Question :

Test the series for convergence or divergence: n=1n51n6+1.\displaystyle\sum_{n=1}^\infty \frac{n^5 - 1}{n^6 + 1}.

Test the series for convergence or divergence:
$\displaystyle\sum_{n=1}^\infty  | Doubtlet.com

Solution:

Neetesh Kumar

Neetesh Kumar | December 23, 2024

Calculus Homework Help

This is the solution to Math 1c
Assignment: 11.7 Question Number 1
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Step-by-step solution:

Step 1: Analyze the general term of the series:

The general term of the series is:

an=n51n6+1.a_n = \frac{n^5 - 1}{n^6 + 1}.

As nn becomes large, the highest powers of nn dominate both the numerator and denominator.

To determine the behavior of the series, simplify ana_n for large nn:

ann5n6=1n.a_n \sim \frac{n^5}{n^6} = \frac{1}{n}.

This suggests that ana_n behaves like the harmonic series term 1n\frac{1}{n}, which is divergent.

Step 2: Apply the Limit Comparison Test:

We compare ana_n with the harmonic series 1n\frac{1}{n}, which diverges.

The Limit Comparison Test states that if:

limnanbn=L,\displaystyle\lim_{n \to \infty} \frac{a_n}{b_n} = L,

where 0<L<0 < L < \infty, then both series either converge or diverge together.

Let bn=1nb_n = \frac{1}{n}. Compute the limit:

limnn51n6+11n=limnn51n6+1n.\displaystyle\lim_{n \to \infty} \frac{\frac{n^5 - 1}{n^6 + 1}}{\frac{1}{n}} = \displaystyle\lim_{n \to \infty} \frac{n^5 - 1}{n^6 + 1} \cdot n.

Simplify:

limnn6nn6+1.\displaystyle\lim_{n \to \infty} \frac{n^{6} - n}{n^6 + 1}.

Divide numerator and denominator by n6n^6:

limn11n51+1n6=101+0=1.\displaystyle\lim_{n \to \infty} \frac{1 - \frac{1}{n^5}}{1 + \frac{1}{n^6}} = \frac{1 - 0}{1 + 0} = 1.

Since the limit is L=1L = 1, and LL is finite and positive, the Limit Comparison Test applies.

Step 3: Conclusion:

Since the harmonic series 1n\sum \frac{1}{n} diverges, and the given series behaves similarly for large nn, the series:

n=1n51n6+1\displaystyle\sum_{n=1}^\infty \frac{n^5 - 1}{n^6 + 1}

is divergent.

Final Answer:

The series is: divergent\boxed{\text{divergent}}


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