Solved: "Use the Limit Comparison Test to determine the..."

Chesley

Answered question

2020-11-08

Use the Limit Comparison Test to determine the convergence or divergence of the series.

\sum_{n = 1}^{\infty} \frac{n^{k - 1}}{n^{k} + 1}, k > 2

Macsen Nixon

Skilled2020-11-09Added 117 answers

Given:

\sum_{n = 1}^{\infty} \frac{n^{k - 1}}{n^{k} + 1}, k > 2

Explanation:
Let

a_{n} = \frac{n^{k - 1}}{n^{k} + 1}

then

a_{n} > 0 \forall n \geq 1

Let

b_{n} = \frac{1}{n}

then

b_{n} > 0

\sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} \frac{1}{n}

is a harmonic series, so it is divergent
Now,

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = lim_{n \to \infty} \frac{\frac{n^{k - 1}}{n^{k} + 1}}{\frac{1}{n}}

= lim_{n \to \infty} \frac{n^{k}}{n^{k} + 1}

= lim_{n \to \infty} \frac{1}{1 + \frac{1}{n^{k}}}

= \frac{1}{1 + 0}

= 1

Since

\sum_{n = 1}^{\infty} \frac{1}{n}

is divergent

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