Show LHS = RHS for this sum (to n-- infinity)?

Question

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Any help would be appreciated

JOHN · Accepted Answer

Let S(x)=tanx/2 + (1/2)tanx/2^2 + (1/2^2)tanx/2^3 +...
∫s(x)dx=2lnsec(x/2) + 2lnsec(x/2^2) + 2lnsec(2/2^3)+..
=2[ln1/{cos(x/2)cos(x/2^2)...to ∞}=2ln(x/sinx)=2lnx - 2ln(sinx)
 [I have proved in Yahoo Answers about a week ago that 
cos(x/2)cos(x/2^2)cos(x/2^3)...cos(x/2^2)..to infinity is sinx/x.
[See http://answers.yahoo.com/question/index;_ylt=ArPFVlu7.svn9YFAeLF20uvty6IX;_ylv=3?qid=20120729081651AAwc43u]
=∫(2/x)dx - ∫2cotx)dx. Hence S(x)=2/x-2cotx.
[I should introduce an arbitrary constant of integration, C, in line 2 above, and then show that C=0. Let x→0.Then the LHS→0 and the RHS→ln1 (since x/sinx→1) =0. Thus C=0].
The 2nd part of the question immediately follows on setting x=π/2.

Show LHS = RHS for this sum (to n--> infinity)?

1 Answer