20110417, 17:32  #1 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3·419 Posts 
Infinite series
Consider the following infinite series that I faced
Prove that its value is equal to ln 2. Leaving away with that first term alone (that is 1/2), the general form for that remaining terms by using partial fractions is being given by which is So, the sum of infinite series is (1/2) + (1/2 + 1/6 + 1/10 + 1/14 + ...)  (1/2 + 1/4 + 1/6 + 1/8 + ...) + (1/6 + 1/10 + 1/14 + 1/18 + ...) = (1/2 + 1/6 + 1/10 + 1/14 + ...) + [1/2 + 1/6 + 1/10 + 1/14 + ...]  [1/2 + 1/4 + 1/6 + 1/8 + ...] Deleting repeated terms within third series from the second series we get that value to be = (1/2 + 1/6 + 1/10 + 1/14 + ...)  (1/4 + 1/8 + 1/12 + 1/16 + ...) = 1/2 (1  1/2 + 1/3  1/4 + ...) = 1/2 (ln 2) What mistake is that I am doing over here? Alternately consider with that series... (1  1/2 + 1/3  1/4 + ...) Its actual value should be equal to ln 2. Writing it as 1  (1  1/2) + 1/3  (1/2  1/4) + 1/5  (1/3  1/6) + 1/7  (1/4  1/8) + 1/9  (1/5  1/10) + ... Convince that it can be rearranged into 1  1 + 1/2  1/2 + 1/3  1/3 + 1/4  1/4 + 1/5  1/5 + ... = 0 Thus, is it true that value of ln 2 = 0? Extra: Can you provide with an equation, which if solved directly gives the correct answer, if squared, terms combined, and then taken square root  if put gives away with wrong answer; if substituting with giving with its right answer? Actually the usual case is that  that way only it is going so, thus Last fiddled with by Raman on 20110417 at 17:57 
20110417, 17:54  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20110417, 19:42  #3  
"William"
May 2003
New Haven
3·787 Posts 
Quote:
1/2 + (infinity)  (infinity) + (infinity) William 

20110419, 07:47  #4  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
Quote:
(1/2) + (1/2 + 1/6 + 1/10 + 1/14 + ...)  (1/2 + 1/4 + 1/6 + 1/8 + ...) + (1/6 + 1/10 + 1/14 + 1/18 + ...) = 2 [1/2 + 1/6 + 1/10 + 1/14 + ...]  (1/2 + 1/4 + 1/6 + 1/8 + ...) = (1 + 1/3 + 1/5 + 1/7 + 1/9 + ...)  (1/2 + 1/4 + 1/6 + 1/8 + 1/10 + ...) = (1  1/2 + 1/3  1/4 + 1/5  1/6 + 1/7  1/8 + 1/9  1/10 + ...) which is equal to ln 2 by using its definition for this purpose Thus, please note about that error! if and only if . It can be considered to be as a limit if that value of x is exactly equal to 1, thus That is not an absolute convergent series at all, by the way Last fiddled with by Raman on 20110419 at 07:53 

20110419, 11:40  #5 
Mar 2009
2·19 Posts 
Do you accept a computer aided proof?
The k'th partial sum of your series is ln(2) + R(k) with the remainder term R(k) = 1/2/(2*k+1)  1/2*Psi(k+1) + 1/2*Psi(k+3/2) where Psi is the Digamma function. The remainder R(k) goes to zero for k > infinity as R(k) ~ 1/2/k  5/16/k^2 + O(1/k^3) 
20110422, 07:54  #6  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3·419 Posts 
Quote:
= = = = = = By the way, here are other classical mathematical errors: On the other hand, consider with these set of equations within that list: Disclaimer: Note that this is actually meant for fun and not for attacking me with that mistakes at all. Last fiddled with by Raman on 20110422 at 08:09 Reason: up down of r ancy i  zearry caseztuchz ny nuch stz 

20110424, 01:21  #7 
Aug 2002
Buenos Aires, Argentina
2^{3}×13^{2} Posts 
The original poster wants to know the limit for 1 of:
Invoking Mathematica Online Integrator: The constant of integration is zero. The constant of integration is zero again. In order to find the limit for x>1 we have to collect the terms which contain ln(1x): where and The conditions for L'HÃ´pital rule hold so we have to compute the derivatives of g(x) and h(x) and the limit of their quotient for x>1: So the terms which include ln(1x) cancel themselves. The limit of the other terms are readily computed since they do not tend to infinite. The limit is as expected by the original poster. Last fiddled with by alpertron on 20110424 at 01:31 
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