- #1

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Can you please help me

Thank you in advance

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The conclusion of the implication is that if $\lim a_k/b_k=c$ and $0<c<\infty$, then the answer for $\sum b_k$ is the same as the one for $\sum a_k$.f

- #1

- 58

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Can you please help me

Thank you in advance

- #2

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MHB

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Have you tried to apply the limit comparison test or find the limit of the terms as $k\to\infty$?

- #3

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okay I will try to do it from what you guide me and I will let you check it for me.Have you tried to apply the limit comparison test or find the limit of the terms as $k\to\infty$?

- #4

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Have you tried to apply the limit comparison test or find the limit of the terms as $k\to\infty$?

I am struggle at this point sir

Please help me

- #5

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I suggest using $a_k=1/k^2$ and $b_k$ as in the original series. Also use the following rule. If $f(x)=ax^m+\sum_{i=0}^{m-1}a_ix^i$ and $g(x)=bx^n+\sum_{i=0}^{n-1}b_ix^i$, then

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\begin{cases}0,&m<n\\a/b,&m=n.\\\infty,&m>n\end{cases}$$

- #6

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Can you guide me more because I am very confused right now ?

I suggest using $a_k=1/k^2$ and $b_k$ as in the original series. Also use the following rule. If $f(x)=ax^m+\sum_{i=0}^{m-1}a_ix^i$ and $g(x)=bx^n+\sum_{i=0}^{n-1}b_ix^i$, then

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\begin{cases}0,&m<n\\a/b,&m=n.\\\infty,&m>n\end{cases}$$

- #7

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I suggest considering $\sum b_k$ to be the series from the problem statement and $a_k=1/k^2$. It is known that $\sum 1/k^2$ converges. What limit do we have to consider to determine the convergence for $\sum b_k$?

- #8

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note ...

$\dfrac{k}{k+2k^2+3k^3} = \dfrac{1}{1+2k+3k^2} < \dfrac{1}{k^2} \text{ for all } k > 1$

$\displaystyle \implies \sum_{k=1}^\infty \dfrac{1}{1+2k+3k^2} \text{ converges by direct comparison to the known convergent series } \sum_{k=1}^\infty \dfrac{1}{k^2}$

- #9

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So here is the work that I have done recently so how can I determine that it is convergent or divergent ?

I suggest considering $\sum b_k$ to be the series from the problem statement and $a_k=1/k^2$. It is known that $\sum 1/k^2$ converges. What limit do we have to consider to determine the convergence for $\sum b_k$?

Thank you in advance

- #10

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so how can I determine that it is convergent or divergent ?

In fact, $\sum_{k=1}^\infty 1/k^2=\pi^2/6$ (see Wikipedia).It is known that $\sum 1/k^2$ converges.

- #11

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I still do not understand what you are trying to tell me sir.In fact, $\sum_{k=1}^\infty 1/k^2=\pi^2/6$ (see Wikipedia).

- #12

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