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 Register FAQ Search Today's Posts Mark Forums Read 2011-04-17, 17:32 #1 Raman 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 2011-04-17 at 17:57   2011-04-17, 17:54   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts Quote:
 Originally Posted by Raman 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
well for the first one if you are multiplying the denominator numbers they have a common factor of 2 to weed out. ( if not I'm confused), and since it turns the first term into 1/1 and you add more since ln(2)<1 it can't be the solution.   2011-04-17, 19:42   #3
wblipp

"William"
May 2003
New Haven

2,371 Posts Quote:
 Originally Posted by Raman 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 + ...)
Justifying this rearrangement requires the series to be absolutely convergent. Each of the three groupings looks like a harmonic series, so the series is not absolutely convergent. In other words, this particular grouping is

1/2 + (infinity) - (infinity) + (infinity)

William   2011-04-19, 07:47   #4
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

3×419 Posts Quote:
 Originally Posted by Raman 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? - what does it mean? any html tag? I don't think so at all Thus, why is it so?
On the other hand, combining & putting together with that terms, somewhat like this gives with results as follows

(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

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 2011-04-19 at 07:53   2011-04-19, 11:40 #5 Gammatester   Mar 2009 468 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)   2011-04-22, 07:54   #6
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

3·419 Posts Quote:
 Originally Posted by Raman 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
Actually that I meant with something like this

= = = = = =

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 2011-04-22 at 08:09 Reason: up down of r ancy i - zearry caseztuchz ny nuch stz   2011-04-24, 01:21 #7 alpertron   Aug 2002 Buenos Aires, Argentina 5A716 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(1-x): where and The conditions for L'Hôpital 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(1-x) 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 2011-04-24 at 01:31  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Greebley Aliquot Sequences 4 2013-02-06 19:28 clowns789 Miscellaneous Math 3 2005-11-11 01:02 jinydu Puzzles 9 2005-10-24 18:23 LoKI.GuZ Math 10 2004-11-28 03:07 eaS Puzzles 20 2003-09-21 02:48

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