MHB Series Convergence and Divergence I

[ardentmed]

Hey guys,

I have a few quick questions for the problem set I'm working on at the moment:

I'm mostly unsure of my response for b. For a, I just split the series into two parts and added 6+3 to get 9, and thus the series is convergent. For c, I got 3/5 after taking the limit, which is divergent because it does not equal 0.

However, for b, I'm having a hard time calculating r so that I can use the a/(1-r) formula to compute the sum. I'd really appreciate some help guys.

Thanks in advance.

 

[Evgeny.Makarov]

For b, using partial fraction decomposition or otherwise, represent each term as a difference of two expressions. Note that the second expression (with a minus) of each term is the same as the first expression (with a plus) of the next term. Therefore, this is a telescoping series.

 

[Prove It]

Hey guys,

I have a few quick questions for the problem set I'm working on at the moment:

I'm mostly unsure of my response for b. For a, I just split the series into two parts and added 6+3 to get 9, and thus the series is convergent. For c, I got 3/5 after taking the limit, which is divergent because it does not equal 0.

However, for b, I'm having a hard time calculating r so that I can use the a/(1-r) formula to compute the sum. I'd really appreciate some help guys.

Thanks in advance.

The convergence of the second is easy, because for large values of n, $\displaystyle \begin{align*} \sum{ \frac{1}{4n^2 - 1} } \end{align*}$ behaves like $\displaystyle \begin{align*} \sum{ \frac{1}{4n^2} } = \frac{1}{4} \sum{ \frac{1}{n^2} } \end{align*}$, which is a convergent p series.

But as Evgeny said, to evaluate the value of the sum you must use a partial fraction decomposition to turn it into a telescopic series.

 

[Prove It]

Hey guys,

I have a few quick questions for the problem set I'm working on at the moment:

I'm mostly unsure of my response for b. For a, I just split the series into two parts and added 6+3 to get 9, and thus the series is convergent. For c, I got 3/5 after taking the limit, which is divergent because it does not equal 0.

However, for b, I'm having a hard time calculating r so that I can use the a/(1-r) formula to compute the sum. I'd really appreciate some help guys.

Thanks in advance.

For a) I have no idea what you are talking about. Adding what 6 and what 3 to get what 9? What does that have to do with convergence?

Write it as $\displaystyle \begin{align*} \sum{ \left( \frac{3}{2^{n-1}} + \frac{2}{3^{n-1}} \right) } = \sum{ \frac{3}{2^{n - 1}} } + \sum{ \frac{2}{3^{n-1}}} = 3\sum{ \left[ \left( \frac{1}{2} \right) ^{n - 1} \right] } + 2\sum{ \left[ \left( \frac{1}{3} \right) ^{n - 1} \right] } \end{align*}$

each of those is a geometric series with $\displaystyle \begin{align*} r_1 = \frac{1}{2} \end{align*}$ and $\displaystyle \begin{align*} r_2 = \frac{1}{3} \end{align*}$. Since these common ratios are all smaller than 1 in size, each of those geometric series is convergent and so the whole sum is convergent.

Also you can evaluate each geometric series using the formula $\displaystyle \begin{align*} \frac{a}{1 - r} \end{align*}$.Your answer to (c) is correct.

 

[ardentmed]

For a) I have no idea what you are talking about. Adding what 6 and what 3 to get what 9? What does that have to do with convergence?

Write it as $\displaystyle \begin{align*} \sum{ \left( \frac{3}{2^{n-1}} + \frac{2}{3^{n-1}} \right) } = \sum{ \frac{3}{2^{n - 1}} } + \sum{ \frac{2}{3^{n-1}}} = 3\sum{ \left[ \left( \frac{1}{2} \right) ^{n - 1} \right] } + 2\sum{ \left[ \left( \frac{1}{3} \right) ^{n - 1} \right] } \end{align*}$

each of those is a geometric series with $\displaystyle \begin{align*} r_1 = \frac{1}{2} \end{align*}$ and $\displaystyle \begin{align*} r_2 = \frac{1}{3} \end{align*}$. Since these common ratios are all smaller than 1 in size, each of those geometric series is convergent and so the whole sum is convergent.

Also you can evaluate each geometric series using the formula $\displaystyle \begin{align*} \frac{a}{1 - r} \end{align*}$.Your answer to (c) is correct.

Oh, I thought that if I separated the series into two parts, the sum of each should be 3 and 6 respectively, thus making the sum of the series 9 (the summation of the two). Is that incorrect?

Thanks in advance.

Edit: Also, for b, the telescoping series, I took the limit as it approaches infinity after crossing out all of the extraneous values. I ended up getting (-.5) +(-.5 /2n) +(-.5/(2n+1) = -1/2. Does that sound about right?

Thanks again.

 

[Prove It]

Oh, I thought that if I separated the series into two parts, the sum of each should be 3 and 6 respectively, thus making the sum of the series 9 (the summation of the two). Is that incorrect?

Now I understand what you meant. Yes that is correct.

 

[ardentmed]

Now I understand what you meant. Yes that is correct.

Thank you. Also, for b, the telescoping series, I took the limit as it approaches infinity after crossing out all of the extraneous values. I ended up getting (-.5) +(-.5 /2n) +(-.5/(2n+1) = -1/2. Does that sound about right?

Thanks again.

 

[Evgeny.Makarov]

Also, for b, the telescoping series, I took the limit as it approaches infinity after crossing out all of the extraneous values. I ended up getting (-.5) +(-.5 /2n) +(-.5/(2n+1) = -1/2. Does that sound about right?

All terms in the series are positive. How did you get $-1/2$? Also, how did you get the denominator $2n$?

 

[ardentmed]

All terms in the series are positive. How did you get $-1/2$? Also, how did you get the denominator $2n$?

Because when using partial fractions, I obtained A= 0.5 and B= -0.5, so the series is 0.5/(2n+1) - (.5)/(2n-1), so when the terms were crossing out, the first term was crossed out when simplifying, whereas the second term stayed put. Taking the limit thereof gave a negative value.

 

[Evgeny.Makarov]

Because when using partial fractions, I obtained A= 0.5 and B= -0.5, so the series is 0.5/(2n+1) - (.5)/(2n-1)

$\frac{1}{2n+1}$ is smaller than $\frac{1}{2n-1}$, so the difference you wrote is negative. In reality,
\[
\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right).
\]

 

[ardentmed]

$\frac{1}{2n+1}$ is smaller than $\frac{1}{2n-1}$, so the difference you wrote is negative. In reality,
\[
\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right).
\]

Alright, so the sum is 1/2 after taking that into consideration, correct?

 

[Prove It]

Alright, so the sum is 1/2 after taking that into consideration, correct?

Yes, well done :)