Expert Answer to "Use the Limit Comparison Test to prove convergence..."

Burhan Hopper

Answered question

2021-01-10

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

\sum_{n = 1}^{\infty} \frac{e^{n} + n}{e^{2 n} - n^{2}}

Brighton

Skilled2021-01-11Added 103 answers

We will use the Comparison Test for convergence of infinite series.
Given infinite series is:

\sum_{n = 1}^{\infty} \frac{e^{n} + n}{e^{2 n} - n^{2}}

Let us consider

a_{n} = \frac{e^{n} + n}{e^{2 n} - n^{2}}

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

Then we get,

\frac{a_{n}}{b_{n}} = \frac{\frac{e^{n} + n}{e^{2 n} - n^{2}}}{\frac{1}{e^{n}}} = e^{n} (\frac{e^{n} + n}{e^{2 n} - n^{2}}) = \frac{e^{2 n} + \neq^{n}}{e^{2 n} - n^{2}} = \frac{1 + \frac{n}{e^{n}}}{1 - \frac{n^{2}}{e^{2 n}}}

lim_{n \to \infty} (\frac{a_{n}}{b_{n}}) = lim_{n \to \infty} (\frac{1 + \frac{n}{e^{n}}}{1 - \frac{n^{2}}{e^{2 n}}}) = 1

Since we can prove by using Integral Test that the infinite series

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

is convergent,
Therefore the given infinite series

\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} \frac{e^{n} + n}{e^{2 n} - n^{2}}

is also convergent by Comparison Test.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?