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Old 2011-04-17, 17:32   #1
Raman
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Default Infinite series

Consider the following infinite series that I faced
\frac{1}{2} + \frac{1}{1.2.3} + \frac{1}{3.4.5} + \frac{1}{5.6.7} + \frac{1}{7.8.9} + ...
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
\frac{1}{(2n-1)(2n)(2n+1)}
which is
\frac{1}{2(2n-1)} - \frac{1}{2n} + \frac{1}{2(2n+1)}

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 \sqrt{x^2} = x gives away with wrong answer; if substituting with \sqrt{x^2} = -x giving with its right answer?

Actually the usual case is that \sqrt{x^2} = |x| - that way only it is going so, thus

Last fiddled with by Raman on 2011-04-17 at 17:57
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Old 2011-04-17, 17:54   #2
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Quote:
Originally Posted by Raman View Post
Consider the following infinite series that I faced
\frac{1}{2} + \frac{1}{1.2.3} + \frac{1}{3.4.5} + \frac{1}{5.6.7} + \frac{1}{7.8.9}
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
\frac{1}{(2n-1)(2n)(2n+1)}
which is
\frac{1}{2(2n-1)} - \frac{1}{2n} + \frac{1}{2(2n+1)}

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 \sqrt{x^2} = x gives away with wrong answer; if substituting with \sqrt{x^2} = -x giving with its right answer?

Actually the usual case is that \sqrt{x^2} = |x| - 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.
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Old 2011-04-17, 19:42   #3
wblipp
 
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Quote:
Originally Posted by Raman View Post
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
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Old 2011-04-19, 07:47   #4
Raman
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Quote:
Originally Posted by Raman View Post
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?

<snip> - 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
Thus, please note about that error!

ln (1 + x) = \sum_{i=1}^{\infty} (-1)^{i+1}\ \frac{x^i}{i}
if and only if |x| < 1.
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
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Old 2011-04-19, 11:40   #5
Gammatester
 
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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)
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Old 2011-04-22, 07:54   #6
Raman
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Quote:
Originally Posted by Raman View Post
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 \sqrt{x^2} = x gives away with wrong answer; if substituting with \sqrt{x^2} = -x giving with its right answer?

Actually the usual case is that \sqrt{x^2} = |x| - that way only it is going so, thus
Actually that I meant with something like this

\frac{q-p}{\sqrt{pq}} = \sqrt{\frac{q}{p}} - \sqrt{\frac{p}{q}} = \sqrt{({\sqrt{\frac{q}{p}} - \sqrt{\frac{p}{q}}})^2} = \sqrt{\frac{q}{p}+\frac{p}{q}-2} = \sqrt{\frac{p^2+q^2-2pq}{pq}} = \sqrt{\frac{(p-q)^2}{pq}} = \frac{p-q}{\sqrt{pq}}

By the way, here are other classical mathematical errors:

1 = \sqrt{1} = \sqrt{-1 \times -1} = \sqrt{-1} \times \sqrt{-1} = i \times i = -1

On the other hand, consider with these set of equations within that list:
 a^2 - b^2 = (a+b)(a-b)
 a^2 - a^2 = (a+a)(a-a)
 a \strike{(a-a)} = 2a \strike{(a-a)}
 \strike{a} = 2 \strike{a}
 1 = 2

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
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Old 2011-04-24, 01:21   #7
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The original poster wants to know the limit for 1- of:

f(x) = \frac{x^3}{1*2*3} + \frac{x^5}{3*4*5} + \frac{x^7}{5*6*7} + \ldots

f'(x) = \frac{x^2}{1*2} + \frac{x^4}{3*4} + \frac{x^6}{5*6} + \ldots

f''(x) = \frac{x}{1} + \frac{x^3}{3} + \frac{x^5}{5} + \ldots = \oper{arctgh} x

Invoking Mathematica Online Integrator:

f'(x) = \frac{1}{2}\ln(1-x^2) + x \oper{arctgh} x

The constant of integration is zero.

f(x) = \frac{x}{2}\ln(1-x^2) + \frac{x^2}{2} \oper{arctgh} x - \frac{x}{2} + \frac{1}{4}\ln(1-x) - \frac{1}{2}\ln(2(1-x)) + \frac{1}{2}\ln(2(1+x)) - \frac{1}{4}\ln(x+1)

The constant of integration is zero again.

f(x) = \frac{x}{2}\ln(1+x) + \frac{x}{2}\ln(1-x) + \frac{x^2}{4} \ln(1+x) - \frac{x^2}{4}\ln(1-x) - \frac{x}{2} + \frac{1}{4}\ln(1-x) - \frac{1}{2}\ln(2) - \frac{1}{2}\ln(1-x) + \frac{1}{2}\ln(2) + \frac{1}{2}\ln(x+1) - \frac{1}{4}\ln(x+1)

In order to find the limit for x->1- we have to collect the terms which contain ln(1-x):

n(x) = (\frac{x}{2}-\frac{x^2}{4}+\frac{1}{4}-\frac{1}{2})\ln(1-x)=\frac{g(x)}{h(x)}

where g(x) = \ln(1-x) and

h(x)=\frac{-4}{x^2-2x+1} = \frac{-4}{(x-1)^2

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-:

\lim_{x\to 1-}{n(x) = \lim_{x\to 1-} \frac{g'(x)}{h'(x)}} = \lim_{x\to 1-} {\frac{1}{x-1}\frac{(x-1)^3}{8}} = \lim_{x\to 1-}{\frac{(x-1)^2}{8}} = 0

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 \ln(2) - \frac{1}{2} as expected by the original poster.

Last fiddled with by alpertron on 2011-04-24 at 01:31
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