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Question Number 9237 by geovane10math last updated on 25/Nov/16

About the Euler-Mascheroni constant γ = lim_(n→∞) (Σ_(k=1) ^n (1/k) − ln n) Why the limit converges if lim_(n→∞) (Σ_(k=1) ^n (1/k) − ln n) = lim_(n→∞) Σ_(k=1) ^n (1/k) − lim_(n→∞) ln n = ∞ − ∞ ?

$${About}\:{the}\:{Euler}-{Mascheroni}\:{constant} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\gamma\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:−\:\mathrm{ln}\:{n}\right) \\ $$$${Why}\:{the}\:{limit}\:{converges}\:{if} \\ $$$$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:−\:\mathrm{ln}\:{n}\right)\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:−\:\underset{{n}\rightarrow\infty} {\mathrm{lim}ln}\:{n} \\ $$$$=\:\infty\:−\:\infty\:? \\ $$

Commented by geovane10math last updated on 25/Nov/16

$${Why}\:\infty\:−\:\infty\:=\:\gamma\:? \\ $$

Commented by 123456 last updated on 25/Nov/16

this limit is idertermined, consider lim_(x→∞) x+(−x) lim_(x→∞) x=∞ lim_(x→∞) −x=−∞ lim_(x→∞) x+(−x)=∞−∞=? =lim_(x→∞) 0=0 but can′t conclude that ∞−∞=0, consider lim_(x→∞) x+(−x+5) lim_(x→∞) −x+5=−∞ lim_(x→∞) x+(−x+5)=∞−∞=? =lim_(x→∞) 5=5 ⇏∞−∞=5 by similiar way (k∈R) lim_(x→∞) x+(−x+k)=∞−∞ =k⇏∞−∞=k so it depends of function nature, and do not always diverge.

$$\mathrm{this}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{idertermined},\:\mathrm{consider} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\:{x}+\left(−{x}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}=\infty \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}−{x}=−\infty \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}+\left(−{x}\right)=\infty−\infty=? \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}0}=\mathrm{0} \\ $$$$\mathrm{but}\:\mathrm{can}'\mathrm{t}\:\mathrm{conclude}\:\mathrm{that}\:\infty−\infty=\mathrm{0}, \\ $$$$\mathrm{consider} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}+\left(−{x}+\mathrm{5}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}−{x}+\mathrm{5}=−\infty \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}+\left(−{x}+\mathrm{5}\right)=\infty−\infty=? \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}5}=\mathrm{5}\:\nRightarrow\infty−\infty=\mathrm{5} \\ $$$$\mathrm{by}\:\mathrm{similiar}\:\mathrm{way}\:\left({k}\in\mathbb{R}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}+\left(−{x}+{k}\right)=\infty−\infty \\ $$$$={k}\nRightarrow\infty−\infty={k} \\ $$$$\mathrm{so}\:\mathrm{it}\:\mathrm{depends}\:\mathrm{of}\:\mathrm{function}\:\mathrm{nature},\:\mathrm{and}\:\mathrm{do} \\ $$$$\mathrm{not}\:\mathrm{always}\:\mathrm{diverge}. \\ $$

Commented by geovane10math last updated on 25/Nov/16

$${Thanks}! \\ $$

Commented by prakash jain last updated on 25/Nov/16

The below statement is true if both limits exist (finite values) lim_(x→a) [f(x)−g(x)]=lim_(x→a) f(x)−lim_(x→a) g(x)

$$\mathrm{The}\:\mathrm{below}\:\mathrm{statement}\:\mathrm{is}\:\mathrm{true}\: \\ $$$$\mathrm{if}\:\mathrm{both}\:\mathrm{limits}\:\mathrm{exist}\:\left(\mathrm{finite}\:\mathrm{values}\right) \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\left[{f}\left({x}\right)−{g}\left({x}\right)\right]=\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)−\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right) \\ $$

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