prove that H_n =Σ_(k=1) ^n (((−1)^(k+1) )/k)×C_n ^k H_n =Σ


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Question Number 73043 by mathmax by abdo last updated on 05/Nov/19

$p r o v e t h a t H_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k + 1}}{k} \times C_{n}^{k}$ $H_{n} = \sum_{k = 1}^{n} \frac{1}{k}$
Commented by mind is power last updated on 06/Nov/19

$B (1, n) = \frac{1}{n} mad this in my anwer . i put \frac{1}{n + 1} but i took \frac{1}{n}$ $i {csn}^{'} t change is pictur that i saved$
Commented by mathmax by abdo last updated on 06/Nov/19
$let p(x) =Σ_(k=1) ^n (((−1)^(k−1) )/k)C_n ^k x^k ⇒ p^′ (x)=Σ_(k=1) ^n (((−1)^(k−1) )/k) C_n ^k kx^(k−1) =Σ_(k=1) ^n C_n ^k (−x)^(k−1) =−(1/x)Σ_(k=1) ^n C_n ^k (−x)^k =−(1/x){(1−x)^n −1} =((1−(1−x)^n )/x) ⇒ p(x) =∫_0 ^x ((1−(1−t)^n )/t)dt +c (c=p(0)=0) ⇒ p(x)=∫_0 ^x ((1−(1−t)^n )/t)dt ⇒p(1)=Σ_(k=1) ^n (((−1)^(k−1) )/k)C_n ^k =∫_0 ^1 ((1−(1−t)^n )/t)dt =_(1−t =u) −∫_0 ^1 ((1−u^n )/(1−u))(−du) =∫_0 ^1 ((1−u^n )/(1−u))du =∫_0 ^1 (((1−u)(1+u+u^2 +...+u^(n−1) ))/(1−u))du =∫_0 ^1 (Σ_(k=0) ^(n−1) u^k )du =Σ_(k=0) ^(n−1) ∫_0 ^1 u^k du =Σ_(k=0) ^(n−1) (1/(k+1)) =Σ_(k=1) ^n (1/k) =H_n so H_n =Σ_(k=1) ^n (((−1)^(k−1) )/k) C_n ^k$
$l e t p (x) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} C_{n}^{k} x^{k} \Rightarrow$ $p^{^{'}} (x) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} C_{n}^{k} {k x}^{k - 1} = \sum_{k = 1}^{n} C_{n}^{k} {(- x)}^{k - 1}$ $= - \frac{1}{x} \sum_{k = 1}^{n} C_{n}^{k} {(- x)}^{k} = - \frac{1}{x} {{(1 - x)}^{n} - 1} = \frac{1 - {(1 - x)}^{n}}{x} \Rightarrow$ $p (x) = \int_{0}^{x} \frac{1 - {(1 - t)}^{n}}{t} d t + c (c = p (0) = 0) \Rightarrow$ $p (x) = \int_{0}^{x} \frac{1 - {(1 - t)}^{n}}{t} d t \Rightarrow p (1) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} C_{n}^{k} = \int_{0}^{1} \frac{1 - {(1 - t)}^{n}}{t} d t$ $=_{1 - t = u} - \int_{0}^{1} \frac{1 - u^{n}}{1 - u} (- d u) = \int_{0}^{1} \frac{1 - u^{n}}{1 - u} d u$ $= \int_{0}^{1} \frac{(1 - u) (1 + u + u^{2} + . . . + u^{n - 1})}{1 - u} d u$ $= \int_{0}^{1} (\sum_{k = 0}^{n - 1} u^{k}) d u = \sum_{k = 0}^{n - 1} \int_{0}^{1} u^{k} d u = \sum_{k = 0}^{n - 1} \frac{1}{k + 1}$ $= \sum_{k = 1}^{n} \frac{1}{k} = H_{n} s o H_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} C_{n}^{k}$
	Answered by mind is power last updated on 06/Nov/19

	Commented by mathmax by abdo last updated on 06/Nov/19

	$t h a n k y o u s i r f o r t h i s h a r d w o r k .$
	Commented by mind is power last updated on 06/Nov/19

	$withe plesur$
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