Determine whether U_n = nth-root(n+1) - nth-root(n) is convergent or divergent

[MisterMaths]

This is about sequences and series, to be more precise it is about finding their nature which is either divergent or convergent by using different methods (Cauchy, D'Alembert, Reiman...). I study maths in french so somethings might not make sense since they are directly translated by me but after all, maths is one language.

I was to determine whether this sequence is convergent or divergent:

. . . . .\(\displaystyle U_n\, =\, \sqrt[\Large{n}]{\strut n\, +\, 1\,}\, -\, \sqrt[\Large{n}]{\strut n\,}\)

 

[Deleted member 4993]

This is about sequences and series, to be more precise it is about finding their nature which is either divergent or convergent by using different methods (Cauchy, D'Alembert, Reiman...). I study maths in french so somethings might not make sense since they are directly translated by me but after all, maths is one language.

I was to determine whether this sequence is convergent or divergent:

. . . . .\(\displaystyle U_n\, =\, \sqrt[\Large{n}]{\strut n\, +\, 1\,}\, -\, \sqrt[\Large{n}]{\strut n\,}\)

What method/s have you been taught to analyze convergence of sequences?

 

[HallsofIvy]

Seeing the difference of two square roots, my first thought is "conjugate". Multiply by \(\displaystyle \frac{\sqrt{n+1}+ \sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\).

 

[Deleted member 4993]

Seeing the difference of two square roots, my first thought is "conjugate". Multiply by \(\displaystyle \frac{\sqrt{n+1}+ \sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\).

I think OP meant "n th root" - not square root.

 

[HallsofIvy]

Okay, thanks. But same basic idea: \(\displaystyle \left(\sqrt[n]{n+1}- \sqrt[n]{n}\right)\left(\sqrt[n]{(n+1)^{n-1}}+ \sqrt[n]{(n+1)^{n-2}}\sqrt[n]{n}\cdot\cdot\cdot+ \sqrt[n]{n+1}\sqrt[n]{n^{n-1}}\right)= n+ 1- n= 1\)

 

[MisterMaths]

What method/s have you been taught to analyze convergence of sequences?

Cauchy's criterion, D'Alembert's test of convergence,Reiman's harmonic series, using the sum of the given sequence (finding it's limit) if it was initially geometric or arithmetic...

 

[MisterMaths]

Okay, thanks. But same basic idea: \(\displaystyle \left(\sqrt[n]{n+1}- \sqrt[n]{n}\right)\left(\sqrt[n]{(n+1)^{n-1}}+ \sqrt[n]{(n+1)^{n-2}}\sqrt[n]{n}\cdot\cdot\cdot+ \sqrt[n]{n+1}\sqrt[n]{n^{n-1}}\right)= n+ 1- n= 1\)

What exactly did you do there?

 

[stapel]

What exactly did you do there?

Like the helper said: the "same basic idea" as the conjugate of the square roots.