Infinite series

Raman · 2011-04-17, 17:32

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

science_man_88 · 2011-04-17, 17:54

Quote:

Originally Posted by Raman

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.

wblipp · 2011-04-17, 19:42

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

Raman · 2011-04-19, 07:47

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?

<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

Gammatester · 2011-04-19, 11:40

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)

Raman · 2011-04-22, 07:54

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 $\sqrt{x^2} = x$ 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.

alpertron · 2011-04-24, 01:21

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.

2011-04-17, 17:32	#1
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 2351₈ Posts	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

2011-04-24, 01:21	#7
alpertron Aug 2002 Buenos Aires, Argentina 1366₁₀ Posts	The original poster wants to know the limit for 1- of: $f(x) = \frac{x^3}{123} + \frac{x^5}{345} + \frac{x^7}{567} + \ldots$ $f'(x) = \frac{x^2}{12} + \frac{x^4}{34} + \frac{x^6}{56} + \ldots$ $f''(x) = \frac{x}{1} + \frac{x^3}{3} + \frac{x^5}{5} + \ldots = \oper{arctgh} x$ 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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2011-04-19, 11:40	#5
Gammatester 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/(2k+1) - 1/2Psi(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)