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Question Number 97795 by abdomathmax last updated on 09/Jun/20
![calculate Σ_(n=1) ^∞ (((−1)^(n−1) )/([(√n)])) [..] meant the floor](Q97795.png)
$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\left[\sqrt{\mathrm{n}}\right]} \\ $$$$\left[..\right]\:\mathrm{meant}\:\mathrm{the}\:\mathrm{floor} \\ $$
Answered by bobhans last updated on 10/Jun/20
![⌊(√n) ⌋ = m ∈N m ≤(√n) ≤ m+1 m^2 ≤ n ≤ m^2 +2m+1 Σ_(n = 1) ^∞ (((−1)^(n+1) )/(⌊(√n) ⌋)) = Σ_(m =1) ^∞ Σ_(⌊(√n)⌋=m) (((−1)^(n+1) )/m) = −Σ_(m =1) ^∞ (1/m) Σ_(n = m^2 ) ^(m^2 +2m) (−1)^n = −Σ_(m = 1) ^∞ (((−1)^m )/m) = [−Σ_(m = 1) ^∞ (((−x)^m )/m) ]_( x=0) ^(x=1) = −Σ_(k = 0) ^∞ (((−x)^(k+1) )/(k+1)) = ∫_0 ^1 (dx/(1+x)) =[ ln(x+1) ]_0 ^1 = ln (2)](Q97829.png)
$$\lfloor\sqrt{\mathrm{n}}\:\rfloor\:=\:\mathrm{m}\:\in\mathbb{N} \\ $$$$\mathrm{m}\:\leqslant\sqrt{\mathrm{n}}\:\leqslant\:\mathrm{m}+\mathrm{1} \\ $$$$\mathrm{m}^{\mathrm{2}} \:\leqslant\:\mathrm{n}\:\leqslant\:\mathrm{m}^{\mathrm{2}} +\mathrm{2m}+\mathrm{1} \\ $$$$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\lfloor\sqrt{\mathrm{n}}\:\rfloor}\:=\:\underset{\mathrm{m}\:=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\lfloor\sqrt{\mathrm{n}}\rfloor=\mathrm{m}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{m}} \\ $$$$=\:−\underset{\mathrm{m}\:=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{m}}\:\underset{\mathrm{n}\:=\:\mathrm{m}^{\mathrm{2}} } {\overset{\mathrm{m}^{\mathrm{2}} +\mathrm{2m}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \\ $$$$=\:−\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{m}} }{\mathrm{m}}\:=\:\left[−\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{x}\right)^{\mathrm{m}} }{\mathrm{m}}\:\right]_{\:\mathrm{x}=\mathrm{0}} ^{\mathrm{x}=\mathrm{1}} \\ $$$$=\:−\underset{\mathrm{k}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{x}\right)^{\mathrm{k}+\mathrm{1}} }{\mathrm{k}+\mathrm{1}}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}}\:=\left[\:\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)\:\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\mathrm{ln}\:\left(\mathrm{2}\right)\: \\ $$
Commented by mathmax by abdo last updated on 11/Jun/20
$$\mathrm{thanks}\:\mathrm{sir}\:\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct} \\ $$