How do you determine the convergence or divergence of #Sigma ((-1)^(n+1)n)/(2n-1)# from #[1,oo)#?

Question

Isabella Ackerman · Accepted Answer

The series:
#sum_(n=1)^oo (-1)^(n+1)n/(2n-1)#
is divergent.

You can determine whether an alternating series converges using Leibniz' criteria, which states that: #sum_(n=1)^oo (-1)^na_n# converges if: (i) #a_n>a_(n+1)# (ii) #lim_n a_n = 0# As the general term of the series above can be expressed as: #a_n = -n/(2n-1)# We can quickly see that: #lim_n a_n = lim_n-n/(2n-1)= lim_n-1/(2-1/n)=-1/2# so that condition (ii) is not met and the series is divergent.

Charles Arocha · Answer

undefined To determine the convergence or divergence of the series $ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{2n-1} $, you can use the Alternating Series Test.

First, check if the terms of the series decrease in absolute value as $ n $ increases. In this series, the terms are $ \frac{n}{2n-1} $, which do decrease as $ n $ increases.

Next, check if the limit of the absolute value of the terms approaches zero as $ n $ approaches infinity. In this case, $ \lim_{n \to \infty} \left| \frac{n}{2n-1} \right| = \frac{1}{2} $, which is not zero.

Since both conditions of the Alternating Series Test are satisfied (terms decrease and limit approaches zero), the series $ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{2n-1} $ converges.