- #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
- Start date
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.
1. What is the best way to approach a difficult question?
The best way to approach a difficult question is to break it down into smaller, more manageable parts. Start by identifying the key components of the question and then brainstorm possible solutions or approaches. It can also be helpful to seek advice from others or do some research to gain a better understanding of the question.
2. How can I improve my critical thinking skills when struggling with a question?
One way to improve critical thinking skills is to practice active listening and questioning. This involves actively engaging with the question and considering different perspectives and possible solutions. It can also be helpful to think critically about the information and evidence presented and to evaluate its credibility and relevance.
3. What should I do if I am still struggling with a question after trying different approaches?
If you are still struggling with a question, try taking a break and coming back to it with a fresh perspective. Sometimes, stepping away and focusing on something else can help you think more creatively and come up with new ideas. You can also seek help from a mentor, teacher, or colleague who may offer a different perspective or approach.
4. How can I overcome feeling overwhelmed when faced with a difficult question?
Feeling overwhelmed is a common reaction when faced with a difficult question. To overcome this, try breaking the question down into smaller parts and tackling them one at a time. It can also be helpful to practice mindfulness and focus on the present moment, rather than getting caught up in worrying about the question.
5. Is it okay to ask for help when struggling with a question?
Yes, it is absolutely okay to ask for help when struggling with a question. In fact, seeking help from others can often lead to new insights and solutions. It is important to remember that asking for help is a sign of strength, not weakness, and can ultimately lead to a better understanding of the question and its possible solutions.
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