Question Number 100785 by 175 last updated on 28/Jun/20
Answered by john santu last updated on 28/Jun/20
![consider m ≤ (( n))^(1/(3 )) ≤ m+1 m^3 ≤ n ≤m^3 +3m^2 +1 Σ_(n = 1) ^∞ (((−1)^(n+1) )/(⌊ (√n) ⌋ )) = Σ_(m=1) ^∞ Σ_(⌊ (√( n)) ⌋=m) (((−1)^(n+1) )/m) = −Σ_(m = 1) ^∞ (((−1)^m )/m) = −[Σ_(m=1) ^∞ (((−x)^m )/m) ]_(x=0) ^(x=1) = ∫_0 ^1 Σ_(k=0) ^∞ (−x)^k dx = ∫_0 ^1 (dx/(1+x)) = [ ln(x+1) ]_0 ^1 = ln(2)](https://www.tinkutara.com/question/Q100787.png)
$$\mathrm{consider}\:\mathrm{m}\:\leqslant\:\sqrt[{\mathrm{3}\:}]{\:\mathrm{n}}\:\leqslant\:\mathrm{m}+\mathrm{1} \\ $$$$\mathrm{m}^{\mathrm{3}} \:\leqslant\:\mathrm{n}\:\leqslant\mathrm{m}^{\mathrm{3}} +\mathrm{3m}^{\mathrm{2}} +\mathrm{1}\: \\ $$$$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\lfloor\:\sqrt{\mathrm{n}}\:\rfloor\:}\:=\:\underset{\mathrm{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\lfloor\:\sqrt{\:\mathrm{n}}\:\rfloor=\mathrm{m}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{m}} \\ $$$$=\:−\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{m}} }{\mathrm{m}}\:=\:−\left[\underset{\mathrm{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{x}\right)^{\mathrm{m}} }{\mathrm{m}}\:\right]_{\mathrm{x}=\mathrm{0}} ^{\mathrm{x}=\mathrm{1}} \\ $$$$=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\mathrm{x}\right)^{\mathrm{k}} \:\mathrm{dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}} \\ $$$$=\:\left[\:\mathrm{ln}\left(\mathrm{x}+\mathrm{1}\right)\:\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:=\:\mathrm{ln}\left(\mathrm{2}\right)\: \\ $$