Question-227139

Question Number 227139 by Spillover last updated on 02/Jan/26

Answered by MrAjder last updated on 03/Jan/26

Σ_(n=1) ^∞ (1/(2n(2n−1)))=^([1]) Σ_(n=1) ^∞ ((1/(2n−1))−(1/(2n))) =^([1]) Σ_(n=1) ^∞ ∫_0 ^1 (x^(2n−2) −x^(2n−1) )dx =∫_0 ^1 (1−x)Σ_(n=1) ^∞ x^(2n−2) dx =∫_0 ^1 (1−x)Σ_(n=0) ^∞ x^(2n) dx =∫_0 ^1 ((1−x)/(1−x^2 ))dx =∫_0 ^1 (dx/(1+x)) =ln(1+x)∣_(0 ) ^1 =ln 2 [1]:(1/k)=∫_(0 ) ^1 x^(k−1) dx ∀k∈N^+

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}\overset{\left[\mathrm{1}\right]} {=}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$\overset{\left[\mathrm{1}\right]} {=}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{2}{n}−\mathrm{2}} −{x}^{\mathrm{2}{n}−\mathrm{1}} \right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}−\mathrm{2}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}} \\ $$$$=\mathrm{ln}\left(\mathrm{1}+{x}\right)\mid_{\mathrm{0}\:} ^{\mathrm{1}} \\ $$$$=\mathrm{ln}\:\mathrm{2} \\ $$$$\left[\mathrm{1}\right]:\frac{\mathrm{1}}{{k}}=\int_{\mathrm{0}\:} ^{\mathrm{1}} {x}^{{k}−\mathrm{1}} {dx}\:\forall{k}\in\mathbb{N}^{+} \\ $$

Commented by Spillover last updated on 03/Jan/26