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Question Number 145025 by mathdanisur last updated on 01/Jul/21

\underset{n = 1}{\sum^{\infty}} \frac{1}{(n + 1) \sqrt{n} + n \sqrt{n + 1}} = ?

Answered by phally last updated on 01/Jul/21

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

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

= \underset{n = 1}{\sum^{\infty}} \frac{(n + 1) \sqrt{n} - n \sqrt{n + 1}}{n^{3} + {2 n}^{2} + n - n^{3} - n^{2}}

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

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

= 1 - \frac{1}{\sqrt{n + 1}}

Commented by mathdanisur last updated on 01/Jul/21

t h a n k y o u S i r, a n s w e r ?

Commented by JDamian last updated on 01/Jul/21

you only have to get now the limit when n --> inf

Answered by mathmax by abdo last updated on 01/Jul/21

S_{n} = \sum_{k = 1}^{n} \frac{1}{(k + 1) \sqrt{k} + k \sqrt{k + 1}} \Rightarrow

S_{n} = \sum_{k = 1}^{n} \frac{(k + 1) \sqrt{k} - k \sqrt{k + 1}}{{(k + 1)}^{2} k - k^{2} (k + 1)}

= \sum_{k = 1}^{n} \frac{(k + 1) \sqrt{k} - k \sqrt{k + 1}}{(k^{2} + 2 k + 1) k - k^{3} - k^{2}}

= \sum_{k = 1}^{n} \frac{(k + 1) \sqrt{k} - k \sqrt{k + 1}}{k^{3} + {2 k}^{2} + k - k^{3} - k^{2}} = \sum_{k = 1}^{n} \frac{(k + 1) \sqrt{k} - k \sqrt{k + 1}}{k (k + 1)}

= \sum_{k = 1}^{n} (\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}}) = 1 - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \dots . . + \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}}

= 1 - \frac{1}{\sqrt{n + 1}} \Rightarrow \lim_{n \to + \infty} S_{n} = 1 = \sum_{n = 1}^{\infty} (\dots .)

Commented by mathdanisur last updated on 01/Jul/21

t h a n k s S i r, a n s w e r 1 - \frac{1}{\sqrt{n + 1}} ?

Commented by mathmax by abdo last updated on 01/Jul/21

no answer is 1

Commented by mathdanisur last updated on 01/Jul/21

t h a n k s S i r