# Thread: discuss the convergence/divergence of (annoying function)

1. ## discuss the convergence/divergence of (annoying function)

i simply cant get the the limit comparison test to work on this following function,

Summation of (from 2 to infinity)[ln(k)^(5/2)/k^(15/8)]

2. $\forall x > 0,\,\ln (x) < x \Rightarrow \quad \ln \left( {k^{\frac{5}{4}} } \right) < k^{\frac{5}{4}} \Rightarrow \quad \frac{5}{4}\ln (k) < k^{\frac{5}{4}}$

$\frac{{\left[ {\ln (k)} \right]^{\frac{5}{2}} }}{{k^{\frac{{15}}{8}} }} < \frac{{\left( {\frac{4}{5}} \right)^{\frac{2}{5}} k^{\frac{1}{2}} }}{{k^{\frac{{15}}{8}} }} = \frac{{\left( {\frac{4}{5}} \right)^{\frac{2}{5}} }}{{k^{\frac{{11}}{8}} }}
$

3. so you're saying that the series converges?? but your b_n is greater than a_n and b_n is convergent by p series equal to 11/8...

how did u solve it, plz explain more

4. Originally Posted by ramzouzy
so you're saying that the series converges?? but your b_n is greater than a_n and b_n is convergent by p series equal to 11/8...
Number one: I did not use limit comparison.
I used simple comparison which you can adapt to limit comparison.
I am no sure that you follow the simple arithmetic involved here.

5. i am still frustrated... i didnt see how you got from step 3 to step 4 and how did u get $
\frac{{\left[ {\ln (k)} \right]^{\frac{5}{2}} }}{{k^{\frac{{15}}{8}} }} < \frac{{\left( {\frac{4}{5}} \right)^{\frac{2}{5}} k^{\frac{1}{2}} }}{{k^{\frac{{15}}{8}} }} = \frac{{\left( {\frac{4}{5}} \right)^{\frac{2}{5}} }}{{k^{\frac{{11}}{8}} }}
$

6. O.K.
Start over. Consider this: $\left[ {\ln (k)} \right]^{\frac{5}{2}} = \left[ {10\ln \left( {k^{\frac{1}{{10}}} } \right)} \right]^{\frac{5}{2}} < \left( {10} \right)^{\frac{5}{2}} \left( {k^{\frac{1}{{10}}} } \right)^{\frac{5}{2}} = \left( {10} \right)^{\frac{5}{2}} \left( {k^{\frac{1}{4}} } \right)$