We have #X_n=sum_(k=1)^n1/(2k-1)-1/2logn# #lim_(n->oo)X_n=#?

1 Answer
Oct 24, 2017

#gamma/2+log2#

Explanation:

Calling #gamma_n = sum_(k=1)^n 1/k - log n# we have

#gamma = lim_(n->oo) sum_(k=1)^n 1/k - log n# the so called Euler-Mascheroni constant.

https://es.wikipedia.org/wiki/Constante_de_Euler-Mascheroni

Now #sum_(n=1)^(2n) 1/k = sum_(k=1)^n 1/(2k) + sum_(k=1)^n 1/(2k-1)# then

#sum_(k=1)^n1/(2k-1) = sum_(k=1)^(2n) 1/k - sum_(k=1)^n 1/(2k)# or

#sum_(k=1)^n 1/(2k-1) = sum_(k=1)^(2n) 1/k-1/2 sum_(k=1)^n 1/k# or

#sum_(k=1)^n 1/(2k-1) =gamma_(2n)+log(2n)-1/2(gamma_n+logn) = gamma_(2n)-1/2gamma_n+log2+logn-1/2logn# and then

#sum_(k=1)^n 1/(2k-1) - 1/2log n = gamma_(2n)-1/2 gamma_n+log2#

and finally

#lim_(n->oo)sum_(k=1)^n 1/(2k-1) - 1/2log n = 1/2gamma+log2#