convergence / divergence help

[shelly89]

\(\displaystyle a_{n} = \dfrac{n^{3}}{1+ln(n)} \)


a) for all n ∈ N. Say whether this sequence is divergent, convergent, and/or bounded.

i think this sequence is divergent, just from looking at it?

b) Give a proof of your answer to part (b). Standard properties of ln n may be assumed


I need help with part 'b', dont know how to start.



thank you,

 

[pka]

\(\displaystyle a_{n} = \dfrac{n^{3}}{1+\ln (n)} \)

a) for all n ∈ N. Say whether this sequence is divergent, convergent, and/or bounded.
i think this sequence is divergent, just from looking at it?

b) Give a proof of your answer to part (b). Standard properties of ln n may be assumed

LaTeX fixed.

\(\displaystyle a_{n} = \dfrac{n^{3}}{1+\ln (n)}>\dfrac{n^{3}}{2\ln (n)} \) for \(\displaystyle n\ge 3 \).

 

[shelly89]

LaTeX fixed.

\(\displaystyle a_{n} = \dfrac{n^{3}}{1+\ln (n)}>\dfrac{n^{3}}{2\ln (n)} \) for \(\displaystyle n\ge 3 \).

could you explain how you got this ? Is there a general rule to show that a sequence converges/ diverges ?


thank you

 

[stapel]

\(\displaystyle a_{n} = \dfrac{n^{3}}{1+\ln (n)}>\dfrac{n^{3}}{2\ln (n)} \) for \(\displaystyle n\ge 3 \).

could you explain how you got this?

For n > 3, is ln(n) less than or greater than 1? Then is ln(n) + 1 less than or greater than ln(n) + ln(n) = 2*ln(n)? If the denominator of a fraction gets bigger, what happens to the value of the entire fraction? What then can you say about 1/(1 + ln(n)) and 1/(2*ln(n))?