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Question Number 202490 by Calculusboy last updated on 27/Dec/23

Commented by aleks041103 last updated on 27/Dec/23
$is {x} the whole part or the fractional part?$
$i s {x} t h e w h o l e p a r t o r t h e f r a c t i o n a l p a r t ?$
Commented by Calculusboy last updated on 28/Dec/23
$the fractional part$
$t h e f r a c t i o n a l p a r t$
	Answered by namphamduc last updated on 28/Dec/23

	$I f y o u t o o k t h i s f r o m F B, y o u k n e w I s o l v e d i t, t h e r e s u l t i s γ (E u l e r M a s c h e r o n i c o n s t a n t) .$
	Answered by witcher3 last updated on 28/Dec/23
	$let(1/x)=t =∫_1 ^∞ {t}.((1/(t−1))).(dt/t^2 )=S =∫_1 ^∞ (({t})/(t^2 (t−1)))dt =∫_1 ^∞ ((t−[t])/(t^2 (t−1)))dt =Σ_(k≥1) ∫_k ^(k+1) ((t−k)/(t^2 (t−1)))dt=Σ_(k≥1) ∫_k ^(k+1) ((1−k)/(t−1))+(k/t^2 )+((k−1)/t)dt =Σ_(k≥1) [(1−k)(ln(k)−ln(k−1)−(k/(k+1))+1)+(k−1)ln(k+1)−(k−1)ln(k) =Σ_(k≥1) ^N [2(1−k)ln(k)−(1−k)ln(k^2 −1)+(1/(k+1)) =Σ_(k=1) ^N {(1−k)ln(k)+kln(k+1)+(k−1)ln(k−1)−kln((k) +ln(k)−ln(1+k)+(1/(k+1))} =Nln(N+1)−ln(2)−Nln(N)+ln(1)−ln(N+1)+ln(2)+H_(N+1) −1 S_N =Nln(1+(1/N))+H_(N+1) −ln(N+1)−1 S=lim_(N→∞) S_N =lim_(N→∞) (Nln(1+(1/N))−1)+lim_(N→∞) (H_(N+1) −ln(N+1)} =γ$
	$let \frac{1}{x} = t$ $= \int_{1}^{\infty} {t} . (\frac{1}{t - 1}) . \frac{dt}{t^{2}} = S$ $= \int_{1}^{\infty} \frac{{t}}{t^{2} (t - 1)} dt$ $= \int_{1}^{\infty} \frac{t - [t]}{t^{2} (t - 1)} dt$ $= \sum_{k ⩾ 1} \int_{k}^{k + 1} \frac{t - k}{t^{2} (t - 1)} dt = \sum_{k ⩾ 1} \int_{k}^{k + 1} \frac{1 - k}{t - 1} + \frac{k}{t^{2}} + \frac{k - 1}{t} dt$ $= \sum_{k ⩾ 1} [(1 - k) (\ln (k) - \ln (k - 1) - \frac{k}{k + 1} + 1) + (k - 1) \ln (k + 1) - (k - 1) \ln (k)$ $= \underset{k ⩾ 1}{\sum^{N}} [2 (1 - k) \ln (k) - (1 - k) \ln (k^{2} - 1) + \frac{1}{k + 1}$ $= \underset{k = 1}{\sum^{N}} {(1 - k) \ln (k) + kln (k + 1) + (k - 1) \ln (k - 1) - kln ((k)$ $+ \ln (k) - \ln (1 + k) + \frac{1}{k + 1}}$ $= Nln (N + 1) - \ln (2) - Nln (N) + \ln (1) - \ln (N + 1) + \ln (2) + H_{N + 1} - 1$ $S_{N} = Nln (1 + \frac{1}{N}) + H_{N + 1} - \ln (N + 1) - 1$ $Double subscripts: use braces to clarify$ $= γ$
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