How Can I Tackle Difficult Questions Effectively?

[jaychay]

Can you please help me
Thank you in advance
I am really struggle with this questions

 

[Prove It]

Test for divergence:

$\displaystyle \begin{align*} \lim_{k \to \infty}{ \left( \frac{k}{3\,k^{\frac{4}{3}} - 1} \right) } &= \lim_{k \to \infty}{ \left( \frac{1}{4\,k^{\frac{1}{3}}} \right) } \textrm{ By L'Hospital's Rule} \\
&= 0 \end{align*}$

so it has a CHANCE of converging.

Test for absolute convergence:

$\displaystyle \begin{align*} \sum_{k = 2}^{\infty}{ \left( \frac{k}{3\,k^{\frac{4}{3}} - 1} \right) } &= \sum_{k = 2}^{\infty}{ \left( \frac{1}{3\,k^{\frac{1}{3}} - \frac{1}{k}} \right) } \\ &\sim \sum_{k = 2}^{\infty}{ \left( \frac{1}{3\,k^{\frac{1}{3}}} \right) } \\ &= \frac{1}{3} \sum_{k=2}^{\infty}{ \left( \frac{1}{k^{\frac{1}{3}}} \right) } \end{align*}$

which is known to be a divergent p-series. So the positive term series would also diverge by the limit comparison (you can check the limit yourself).

Test for conditional convergence: Since it's an alternating series, we just need to show that the terms are decreasing.

$\displaystyle \begin{align*} f(x) &= \frac{x}{3\,x^{\frac{4}{3}} - 1} \\
f'\left( x \right) &= -\frac{\left( x^{\frac{4}{3}} + 1 \right) }{\left( 3\,x^{\frac{4}{3}} - 1 \right) ^2 } \end{align*}$

which is very clearly negative for all $\displaystyle \begin{align*} x > 0 \end{align*}$. Thus the terms (which are on the function) also decrease.

Therefore the series is CONDITIONALLY CONVERGENT by the Alternating Series Test.

 

[soloenergy]

and I am not sure where to begin. any guidance would be greatly appreciated.

Hi there,

I would be happy to help you with your questions. Can you provide more information or context about the questions you are struggling with? This way, I can better understand the problem and provide more specific guidance. Looking forward to hearing back from you.