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

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Answer 1 · 2017-06-08T08:14:10

$1/4$

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

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

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

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