How do you use the direct comparison test to determine if #Sigma 1/(n-1)# from #[1,oo)# is convergent or divergent?

Question

Elijah Abram · Accepted Answer

Note that #1/(n-1)>1/n#. You can think about this logically by comparing that #1/2>1/3#, #1/9>1/10#, and so on. Since #sum_(n=2)^oo1/n# is divergent #1/(n-1)>1/n#, we see that #sum_(n=2)^oo1/(n-1)# is divergent as well through the direct comparison test. (Note that the question asks for #sum_(n=1)^oo1/(n-1)#, which is undefined for #n=1#, so I've begun the series at #n=2# instead.)

William Abdalla · Answer

undefined To use the direct comparison test for the series $ \sum \frac{1}{n-1} $ from $ n = 1 $ to infinity, we compare it with the series $ \sum \frac{1}{n} $. Since for all $ n \geq 2 $, $ \frac{1}{n-1} $ is less than or equal to $ \frac{1}{n} $, and $ \sum \frac{1}{n} $ is the harmonic series which diverges, we conclude that $ \sum \frac{1}{n-1} $ also diverges.

Christopher Agresta · Answer

undefined To use the direct comparison test to determine if the series $ \sum_{n=1}^{\infty} \frac{1}{n-1} $ is convergent or divergent, we need to compare it to a known series whose convergence or divergence is already established.

In this case, we can compare the given series to the harmonic series, $ \sum_{n=1}^{\infty} \frac{1}{n} $, which is a well-known divergent series.

To apply the direct comparison test, we need to show that the given series is either less than or greater than the harmonic series for all $ n $ beyond some point.

Consider $ n = 2 $ and beyond:

For $ n \geq 2 $, we have $ \frac{1}{n-1} \geq \frac{1}{n} $, because $ n-1 $ is always less than or equal to $ n $.

Since $ \frac{1}{n-1} \geq \frac{1}{n} $ for all $ n \geq 2 $, and the harmonic series $ \sum_{n=1}^{\infty} \frac{1}{n} $ is known to be divergent, by the direct comparison test, the series $ \sum_{n=1}^{\infty} \frac{1}{n-1} $ is also divergent.