- #1
jaychay
- 58
- 0
Can you please help me
Thank you in advance
I am really struggle with this questions
How Can I Tackle Difficult Questions Effectively?
- MHB
- Thread starter jaychay
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In summary: Thank you. In summary, the conversation is about a series of tests, including testing for divergence, absolute convergence, and conditional convergence, and the usage of L'Hospital's Rule and the Alternating Series Test. The person is asking for guidance and help with their questions and expresses appreciation in advance.
Can you please help me
Thank you in advance
I am really struggle with this questions
- #2
Prove It
Gold Member
MHB
- 1,465
- 24
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.
$\displaystyle \begin{align*} \lim_{k \to \infty}{ \left( \frac{k}{3\,k^{\frac{4}{3}} - 1} \right) } &= \lim_{k \to \infty}{ \left( \frac{1}{4\,k^{\frac{1}{3}}} \right) } \textrm{ By L'Hospital's Rule} \\
&= 0 \end{align*}$
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.
- #3
soloenergy
- 64
- 0
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.
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.
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