# A pair of sequence problems

• Oct 7th 2009, 12:36 PM
matt.qmar
A pair of sequence problems
Does this sequence diverge or converge? If it converges, find the limit-

$\displaystyle (\frac{1}{n})^{(\frac{1}{n})}$

I know the harmonic sequence diverges, but I've taken natural logs of both sides and get stuck, at (1/n)(ln 1 - ln n) = ln L where L is the original limit...

Also,

$\displaystyle \frac{\sqrt{n}}{2n*\csc{\sqrt{9n+4}}}$

can this be shown to converge to 0 using the squeeze theorem for sequence convergence? (note that you can move csc to the numerator as $\displaystyle \sin{\sqrt{9n+4}}$)

(both sequences are from n=1 to infinity for the natural numbers)
• Oct 7th 2009, 01:38 PM
tonio
Quote:

Originally Posted by matt.qmar
Does this sequence diverge or converge? If it converges, find the limit-

$\displaystyle (\frac{1}{n})^{(\frac{1}{n})}$

I know the harmonic sequence diverges, but I've taken natural logs of both sides and get stuck, at (1/n)(ln 1 - ln n) = ln L where L is the original limit...

== I hope you already studied functions and their limits: then what you did is fine, and you need the lim (1/n) lim(1/n) when n--> oo, which can be calculated from what we about lim x ln x when x--> 0 : this limit's value is zero since x --> 0 much faster than ln x --> -oo when x --> 0.
you can do it also using L'Hospital's rule ==. the original limit is 1.

Also,

$\displaystyle \frac{\sqrt{n}}{2n*\csc{\sqrt{9n+4}}}$

can this be shown to converge to 0 using the squeeze theorem for sequence convergence? (note that you can move csc to the numerator as $\displaystyle \sin{\sqrt{9n+4}}$)

If csc is cosecant, the the function is [sin(sqrt(9n+4))]/2]*1/Sqrt(n) and
since 1/Sqrt(n) --> 0 when n --> oo and [sin(sqrt(9n+4))]/2] is bounded the limit is zero.

Tonio

(both sequences are from n=1 to infinity for the natural numbers)

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