How do you test the series #Sigma n/(n+4)# from n is #[0,oo)# for convergence?

Question

Cameron Alexander · Accepted Answer

undefined To test the series $ \sum \frac{n}{n + 4} $ for convergence, you can use the Ratio Test:

1. Compute the limit:
$$ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| $$
where $ a_n = \frac{n}{n + 4} $.

2. Calculate $ a_{n+1} $ by replacing $ n $ with $ n + 1 $ in the formula for $ a_n $.

3. Simplify the ratio $ \frac{a_{n+1}}{a_n} $.

4. Take the limit as $ n $ approaches infinity.

5. If the limit is less than 1, the series $ \sum \frac{n}{n + 4} $ converges. If it's greater than 1 or the limit doesn't exist, the series diverges. If the limit equals 1, the test is inconclusive, and you may need to use another test.

Applying these steps:
$$ a_n = \frac{n}{n + 4} $$
$$ a_{n+1} = \frac{n+1}{n+1+4} = \frac{n+1}{n+5} $$
$$ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} \left| \frac{\frac{n+1}{n+5}}{\frac{n}{n+4}} \right| = \lim_{{n \to \infty}} \left| \frac{n(n+4)}{(n+1)(n+5)} \right| = 1 $$

Since the limit equals 1, the Ratio Test is inconclusive for this series. You may need to use another test, such as the Comparison Test or the Limit Comparison Test, to determine convergence or divergence.

Christopher Allers · Answer

#sum_(n=0)^oo n/(n+4) = oo#

A necessary condition for any series #sum a_n# to converge is that:

#lim_(n->oo) a_n = 0#

In our case:

#lim_(n->oo) n/(n+4) = 1#

so the series is not convergent.

As:

#n/(n+4) > 0#

then the series is divergent.