How do you find this ? #lim_(x->0)sum_(n=1)^oo(cosnx)/((4n-3)(4n+1))#

1 Answer
Cesareo R.
Jun 8, 2017

#1/4#

Explanation:

#lim_(x->0)sum_(n=1)^oo(cosnx)/((4n-3)(4n+1))=sum_(n=1)^oo 1/((4n-3)(4n+1))#

but

# 1/((4n-3)(4n+1)) = 1/4(1/(4n-3)-1/(4n+1))#

and

#sum_(n=1)^m1/(4n-3)-sum_(n=1)^m 1/(4n+1) = 1-1/(4m+1)# so

#lim_(x->0)sum_(n=1)^oo(cosnx)/((4n-3)(4n+1))=1/4#

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