How do you determine the convergence or divergence of #Sigma 1/ncosnpi# from #[1,oo)#?

Question

Scarlett Allison · Accepted Answer

#sum_(n=1)^oo 1/ncospin = -ln2#

First, we note that #cosnpi=(-1)^n# so that the series can be written as:

#sum_(n=1)^oo (-1)^n/n#

This series satisfies Leibniz' criteria as:

#lim_n 1/n = 0# #1/n > 1/(n+1)#

so the series is convergent.

To determine the sum, let's consider a geometric series of ratio #-alpha# with #0 < alpha < 1#.

#sum_(n=0)^oo (-1)^nalpha^n = 1/(1+alpha)#

This series is absolutely convergent and so it can be integrated term by term:

#int_0^x (dalpha)/(1+alpha) = sum_(n=0)^oo int_0^x (-1)^nalpha^ndalpha#

#ln|1+x| = sum_(n=0)^oo (-1)^nx^(n+1)/(n+1)=-sum_(n=0)^oo (-1)^(n+1)x^(n+1)/(n+1)=-sum_(n=1)^oo (-1)^n(x^n)/n#

This equation holds for every #x in (0,1)# and passing to the limit for #x->1# we have:

#ln2 = -sum_(n=1)^oo (-1)^n/n#

Emilia Aguilar · Answer

undefined To determine the convergence or divergence of the series Σ(1/n*cos(nπ)) from n=1 to ∞, you can use the alternating series test. This test states that if a series has alternating signs and the absolute value of the terms decreases monotonically to zero, then the series converges. In this series, the terms alternate in sign due to the cosine function. Additionally, since |cos(nπ)| equals 1 for all n, the terms |1/n*cos(nπ)| decrease monotonically to zero as n increases. Therefore, by the alternating series test, the series Σ(1/n*cos(nπ)) converges.