Prove that Σ_(m=1) ^n (((−1)^m ∙(2^m −1) ∙^n C_m )/m) =Σ


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Question Number 4103 by prakash jain last updated on 28/Dec/15

$Prove that$ $\underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m} \cdot (2^{m} - 1) \cdot^{n} C_{m}}{m} = \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}}{m}$
Commented by Yozzii last updated on 29/Dec/15
$Let z(n)=l.h.s and f(n)=r.h.s. n=1: z(1)=(((−1)(2−1))/1)× ((1),(1) )=−1 f(1)=(((−1)^1 )/1)=−1 ⇒z(1)=f(1). z(2)=z(1)+(((−1)^2 (2^2 −1))/2) ((2),(2) ) z(2)=−1+(3/2) f(2)=f(1)+(((−1)^2 )/2)=−1+(1/2) ⇒z(2)≠f(2). The claim that z(n)=f(n) ∀n∈N is false. If the claim were true, proof is required for (2^m −1) ((n),(m) )=1 ∀m,n∈N+{0}, m≤n, as both sums are over equal indices. ∴ ((n),(m) )=(1/(2^m −1)). (1/(2^m −1))∈Z^+ iff m=1. ⇒ n=1 so that ((n),(m) )= ((1),(1) )=1=(1/(2−1)) Otherwise, (1/(2^m −1))∉Z^+ ⇒ ((n),(m) )≠(1/(2^m −1)) ∀m,n∈N+{0}−{1}.$
$L e t z (n) = l . h . s a n d f (n) = r . h . s .$ $n = 1 : z (1) = \frac{(- 1) (2 - 1)}{1} \times (\begin{matrix} 1 \\ 1 \end{matrix}) = - 1$ $f (1) = \frac{{(- 1)}^{1}}{1} = - 1 \Rightarrow z (1) = f (1) .$ $z (2) = z (1) + \frac{{(- 1)}^{2} (2^{2} - 1)}{2} (\begin{matrix} 2 \\ 2 \end{matrix})$ $z (2) = - 1 + \frac{3}{2}$ $f (2) = f (1) + \frac{{(- 1)}^{2}}{2} = - 1 + \frac{1}{2}$ $\Rightarrow z (2) \neq f (2) .$ $T h e c l a i m t h a t z (n) = f (n) \forall n \in N i s$ $f a l s e .$ $I f t h e c l a i m w e r e t r u e, p r o o f i s$ $r e q u i r e d f o r (2^{m} - 1) (\begin{matrix} n \\ m \end{matrix}) = 1 \forall m, n \in N + {0},$ $m ⩽ n, a s b o t h s u m s a r e o v e r e q u a l i n d i c e s .$ $∴ (\begin{matrix} n \\ m \end{matrix}) = \frac{1}{2^{m} - 1} . \frac{1}{2^{m} - 1} \in Z^{+} i f f m = 1 .$ $\Rightarrow n = 1 s o t h a t (\begin{matrix} n \\ m \end{matrix}) = (\begin{matrix} 1 \\ 1 \end{matrix}) = 1 = \frac{1}{2 - 1}$ $O t h e r w i s e, \frac{1}{2^{m} - 1} \notin Z^{+} \Rightarrow (\begin{matrix} n \\ m \end{matrix}) \neq \frac{1}{2^{m} - 1}$ $\forall m, n \in N + {0} - {1} .$
Commented by prakash jain last updated on 29/Dec/15

$I should have made these basic checks before$ $asking for proof . Yozzi had shown in earlier$ $question Q 2088 .$ $\int_{0}^{π / 4} {t a n}^{2 n + 1} θ d θ = {(- 1)}^{n} (\frac{1}{2} l n 2 + \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}}{2 m})$ $In the same time I did a different integral .$ $\int_{0}^{π / 4} \tan^{2 n} θ \tan θ d θ$ $= \int_{0}^{π / 4} {(\sec^{2} θ - 1)}^{n} \tan θ d θ$ $= {(- 1)}^{n} \int_{0}^{π / 4} {(1 - \sec^{2} θ)}^{n} \tan θ d θ$ $= {(- 1)}^{n} \int_{0}^{π / 4} [1 + \underset{m = 1}{\sum^{n}} {(- 1)}^{m}^{n} C_{m} \sec^{2 m} θ] \tan θ d θ$ $= {(- 1)}^{n} \int_{0}^{π / 4} \tan θ d θ + \int_{0}^{π / 4} \underset{m = 1}{\sum^{n}} [{(- 1)}^{m}^{n} C_{m} \sec^{2 m - 1} θ \cdot \sec θ \tan θ] d θ$ $= {(- 1)}^{n} {[\ln \sec θ + \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}^{n} C_{m} \sec^{2 m} θ}{2 m}]}_{0}^{π / 4}$ $= {(- 1)}^{n} (\ln \sqrt{2} + \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}^{n} C_{m} (2^{m} - 1)}{2 m})$ $Now if I compare the two results it looks like .$ $\underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}^{n} C_{m} (2^{m} - 1)}{2 m} = \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}}{2 m}$ $which is not correct .$ $So where is the error ?$
Commented by Yozzii last updated on 29/Dec/15

$I r e c a l l t h e q u e s t i o n a n d o u r$ $d i s c u s s i o n o n i t . I^{'} l l c h e c k t h e$ $w o r k i n g s o o n .$
Commented by prakash jain last updated on 29/Dec/15

$Thanks . There must be some error in the$ $integral above . I will also recheck .$
Commented by Yozzii last updated on 29/Dec/15

$I t h o u g h t t h a t t h e i d e n t i t y w o r k e d$ $i n d u c t i v e l y b u t a p p a r e n t l y i t d o e s n o t .$ $I n o t h e r w o r d s, y o u c a n n o t a d d t h e n t h$ $t e r m t o t h e p a r t i a l s u m o f n - 1 t e r m s$ $t o g i v e t h e n t h p a r t i a l s u m i n z (n) .$ $F o r n = 2 :$ $z (2) = \frac{{(- 1)}^{1} (2^{1} - 1) (\begin{matrix} 2 \\ 1 \end{matrix})}{1} + \frac{{(- 1)}^{2} (2^{2} - 1) (\begin{matrix} 2 \\ 2 \end{matrix})}{2}$ $z (2) = - 2 + \frac{3}{2} = \frac{- 1}{2} = f (2) = \frac{- 1}{1} + \frac{1}{2}$ $I f d o n e i n d u c t i v e l y, a s d e t a i l e d a b o v e,$ $z (2) = z (1) + \frac{3}{2} = \frac{{(- 1)}^{1} (2 - 1) (\begin{matrix} 1 \\ 1 \end{matrix})}{1} + \frac{3}{2} = \frac{1}{2} \neq f (2)$ $H o w e v e r, t h e s u m \underset{m = 1}{\sum^{n}} \frac{{(- 1)}^{m}}{m} w o r k s$ $i n d u c t i v e l y .$ $S o r r y f o r m y e r r o r i n m y f i r s t c o m m e n t .$
Commented by prakash jain last updated on 29/Dec/15

$Ok . So the assertion in question is not$ $disproved .$ $I also did some computer test for very large$ $n and both LHS and RHS converged to - \ln 2 .$ $Is there a way to prove the statement in$ $question ?$
Commented by Yozzii last updated on 29/Dec/15

$U s i n g t h e i n t e g r a l e q u i v a l e n c e t h a t$ $s u r f a c e d i n Q 2088 ?$
Commented by prakash jain last updated on 29/Dec/15

$That is an option but without using the$ $integral . Suppose the question is asked as I$ $asked now, it is hard to imagine the integral$ $to prove .$ $I have made a few attempts but could not$ $get a simple proof . May by induction ?$
Commented by Yozzii last updated on 29/Dec/15

${T h a t}^{'} s w h a t c a m e t o m i n d . U s e o f$ $(\begin{matrix} n + 1 \\ m \end{matrix}) = (\begin{matrix} n \\ m \end{matrix}) + (\begin{matrix} n \\ m - 1 \end{matrix})$ $m a y a r i s e .$
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