calculate-n-1-1-n-1-n-meant-the-floor-

Question Number 97795 by abdomathmax last updated on 09/Jun/20

calculate \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{[\sqrt{n}]}

[. .] meant the floor

Answered by bobhans last updated on 10/Jun/20

⌊ \sqrt{n} ⌋ = m \in N

m ⩽ \sqrt{n} ⩽ m + 1

m^{2} ⩽ n ⩽ m^{2} + 2 m + 1

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{⌊ \sqrt{n} ⌋} = \underset{m = 1}{\sum^{\infty}} \sum_{⌊ \sqrt{n} ⌋ = m} \frac{{(- 1)}^{n + 1}}{m}

= - \underset{m = 1}{\sum^{\infty}} \frac{1}{m} \underset{n = m^{2}}{\sum^{m^{2} + 2 m}} {(- 1)}^{n}

= - \underset{m = 1}{\sum^{\infty}} \frac{{(- 1)}^{m}}{m} = {[- \underset{m = 1}{\sum^{\infty}} \frac{{(- x)}^{m}}{m}]}_{x = 0}^{x = 1}

= - \underset{k = 0}{\sum^{\infty}} \frac{{(- x)}^{k + 1}}{k + 1} = \underset{0}{\int^{1}} \frac{dx}{1 + x} = {[\ln (x + 1)]}_{0}^{1}

= \ln (2)

Commented by mathmax by abdo last updated on 11/Jun/20

thanks sir your answer is correct