Deep Rooted Tree

Algebra Level 4

Find the minimum value of non-negative integer p p such that k = 1 1 k ( k + 1 ) ( k + 2 ) . . . ( k + p ) \displaystyle \sum_{k=1}^{\infty} \cfrac{1}{\sqrt{k(k+1)(k+2)...(k+p)}} converges.

You can try more of my fundamental problems here .

The series just doesn't converge. 0 3 2 1

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2 solutions

Brian Moehring
Jul 7, 2018

Relevant wiki: Convergence Tests

Note: For reasons that should become obvious, I'm going to use q q instead of p p

For every non-negative integer q q , k = q + 1 1 k ( q + 1 ) / 2 = k = 1 1 ( k + q ) ( q + 1 ) / 2 k = 1 1 k ( k + 1 ) ( k + 2 ) ( k + q ) k = 1 1 k ( q + 1 ) / 2 \sum_{k=q+1}^\infty \frac{1}{k^{(q+1)/2}} = \sum_{k=1}^\infty \frac{1}{(k+q)^{(q+1)/2}} \leq \sum_{k=1}^\infty \frac{1}{\sqrt{k(k+1)(k+2)\cdots (k+q)}} \leq \sum_{k=1}^\infty \frac{1}{k^{(q+1)/2}}

Therefore, the series in question is bounded on both sides by a p p -series with p = q + 1 2 p = \frac{q+1}{2} . It follows that it will converge if and only if q + 1 2 = p > 1 q > 1 \frac{q+1}{2} = p > 1 \iff q > 1

Since q q is an integer, the smallest value is 2 \boxed{2} .

@Brian Moehring Thanks for your solution. I did not realise there is a wiki page on convergence test in Brilliant.

donglin loo - 2 years, 11 months ago
Donglin Loo
Jul 7, 2018

Read this . It explains all.

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