Series Convergence Test Questions

[ardentmed]

View attachment 2803

Just a few quick questions this time:

I'm doubting the first one mostly, because when I used the integral test to evaluate it: I ended up getting (-1/x)(lnx +1) from 2 to infinity, which gave me an odd expression: (-1/infinity)(infinity +1 -ln2 -1). I'm assuming this means it is convergent since it should ultimately equal zero, correct?

As for the b, c, and d, I got convergent by limit comparison, convergent by the ratio test (since r is 0, I'm doubting my answer to this one too.), and convergent by the ratio test.

Thanks.

 

[Prove It]

View attachment 2803

Just a few quick questions this time:

I'm doubting the first one mostly, because when I used the integral test to evaluate it: I ended up getting (-1/x)(lnx +1) from 2 to infinity, which gave me an odd expression: (-1/infinity)(infinity +1 -ln2 -1). I'm assuming this means it is convergent since it should ultimately equal zero, correct?

As for the b, c, and d, I got convergent by limit comparison, convergent by the ratio test (since r is 0, I'm doubting my answer to this one too.), and convergent by the ratio test.

Thanks.

Well first of all, if $\displaystyle \begin{align*} x \geq 2 \end{align*}$ is $\displaystyle \begin{align*} \frac{\ln{(x)}}{x^2} \end{align*}$ increasing or decreasing? What does this tell you about your sum (is it an under-estimate or an over-estimate of your integral?)

 

[ardentmed]

Well first of all, if $\displaystyle \begin{align*} x \geq 2 \end{align*}$ is $\displaystyle \begin{align*} \frac{\ln{(x)}}{x^2} \end{align*}$ increasing or decreasing? What does this tell you about your sum (is it an under-estimate or an over-estimate of your integral?)

If it's increasing ,isn't it divergent?

 

[Prove It]

If it's increasing ,isn't it divergent?

Well yes, so it's a good thing this is a DECREASING function (how could you show that?) but even so, it's not enough to prove convergence. You need to set up either a left-hand estimate or a right-hand estimate of the integral (which will be equal to your sum). One will be an over-estimate, so if you can show the integral is divergent, then the over-estimate sum will also diverge. One will be an under-estimate, so if you can show the integral is convergent, then the under-estimate sum will also converge.

 

[Opalg]

Just a few quick questions this time:

I'm doubting the first one mostly, because when I used the integral test to evaluate it: I ended up getting (-1/x)(lnx +1) from 2 to infinity, which gave me an odd expression: (-1/infinity)(infinity +1 -ln2 -1). I'm assuming this means it is convergent since it should ultimately equal zero, correct?

It looks as though you are using the integral test for this one. So you want to know what happens to \(\displaystyle -\frac{\ln x+1}x\) as $x\to\infty.$ It's no use just putting "$x=\infty$" because that gives an odd expression (as you put it) which in fact is meaningless. What you need to do is to find the limiting value of that fraction as $x$ gets large. It's true that both the numerator $\ln x+1$ and the denominator $x$ will tend to infinity. But one of them goes to infinity much more slowly than the other ... .

As for the b, c, and d, I got convergent by limit comparison, convergent by the ratio test (since r is 0, I'm doubting my answer to this one too.), and convergent by the ratio test.

Correct for b and c (and $r=0$ still gives convergence in the ratio test). For d you need to think again (hint: the terms in d are alternating in sign).

 

[chisigma]

We analyze the series ...

$\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n^{2}}\ (1)$

In my opinion the best way to demonstrate that (1) converges is to use direct comparison test. Setting $\displaystyle a_{n} = \frac{\ln n}{n^{2}}$ and $\displaystyle b_{n} = \frac{1}{n^{\frac{3}{2}}}$ it is easy to demonstrate that is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = \lim_{n \rightarrow \infty} \frac{\ln n}{\sqrt{n}} = 0\ (2)$

... so that for n 'large enough' is $\displaystyle a_{n}<b_{n}$ and since the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}$ converges, the series (1) converges too.

Remembering the definition of the Riemann Zeta Function...

$\displaystyle \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}\ (3)$

... deriving (3) You obtain...

$\displaystyle \zeta^{\ '} (s) = - \sum_{n=1}^{\infty} \frac{\ln n}{n^{s}}\ (4)$

... so that is...

$\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n^{2}} = - \zeta^{\ '} (2) = \frac{\pi^{2}}{6}\ \{12\ \ln A - \gamma - \ln (2\ \pi)\}= .93754825431... (5)$

... where $\gamma$ is the Euler's constant and A the Gleisher-Kinkelin constant...

Kind regards

$\chi$ $\sigma$

 

[ardentmed]

Well yes, so it's a good thing this is a DECREASING function (how could you show that?) but even so, it's not enough to prove convergence. You need to set up either a left-hand estimate or a right-hand estimate of the integral (which will be equal to your sum). One will be an over-estimate, so if you can show the integral is divergent, then the over-estimate sum will also diverge. One will be an under-estimate, so if you can show the integral is convergent, then the under-estimate sum will also converge.

It looks as though you are using the integral test for this one. So you want to know what happens to \(\displaystyle -\frac{\ln x+1}x\) as $x\to\infty.$ It's no use just putting "$x=\infty$" because that gives an odd expression (as you put it) which in fact is meaningless. What you need to do is to find the limiting value of that fraction as $x$ gets large. It's true that both the numerator $\ln x+1$ and the denominator $x$ will tend to infinity. But one of them goes to infinity much more slowly than the other ... .Correct for b and c (and $r=0$ still gives convergence in the ratio test). For d you need to think again (hint: the terms in d are alternating in sign).

Therefore, for a, the series converges because lnx+1 takes much longer than x to reach infinity.

As for d, due to the (-1)^n, the alternating series test must be used, in which case 0 is obtained when taking the limit, since lnn takes longer than n to reach infinity.

Am I on the right track?

Thanks in advance.

 

[Prove It]

Therefore, for a, the series converges because lnx+1 takes much longer than x to reach infinity.

NO! The decreasing nature of the function tells you NOTHING about the convergence of the series, only that it MAY converge.

Assuming we are using a right-hand endpoint approximation of the integral, then we have $\displaystyle \begin{align*} \sum_{n = 2}^{\infty} { \frac{\ln{(n)}}{n^2} } \leq \int_1^{\infty}{ \frac{\ln{(x)}}{x^2}\,\mathrm{d}x } \end{align*}$

You have to now check if that integral is convergent. If it is, then so is the sum.

As for d, due to the (-1)^n, the alternating series test must be used, in which case 0 is obtained when taking the limit, since lnn takes longer than n to reach infinity.

Am I on the right track?

You are partly correct. You are correct that the alternating series test should be used. But again, taking the limit of the terms tells you NOTHING about the convergence of the series, just that they HAVE to go to 0 so that it MAY converge.

For an alternating series to converge, the non-alternating part of the terms have to be DECREASING. How can you show if a function is decreasing?

 

[ardentmed]

NO! The decreasing nature of the function tells you NOTHING about the convergence of the series, only that it MAY converge.

Assuming we are using a right-hand endpoint approximation of the integral, then we have $\displaystyle \begin{align*} \sum_{n = 2}^{\infty} { \frac{\ln{(n)}}{n^2} } \leq \int_1^{\infty}{ \frac{\ln{(x)}}{x^2}\,\mathrm{d}x } \end{align*}$

You have to now check if that integral is convergent. If it is, then so is the sum.
You are partly correct. You are correct that the alternating series test should be used. But again, taking the limit of the terms tells you NOTHING about the convergence of the series, just that they HAVE to go to 0 so that it MAY converge.

For an alternating series to converge, the non-alternating part of the terms have to be DECREASING. How can you show if a function is decreasing?

Perform the integral test on the positive part, right?

 

[Prove It]

Perform the integral test on the positive part, right?

OK good point, the function isn't decreasing over that whole integral. really you need to say $\displaystyle \begin{align*} \sum_{n = 3}^{\infty}{ \frac{ \ln{(n)}}{n^2} } \leq \int_2^{\infty} { \frac{\ln{(x)}}{x^2}\,\mathrm{d}x } \end{align*}$ as it is in this region that the entire function is decreasing.

Showing that this integral is convergent (i.e. evaluating the integral) will show that this particular series converges. Adding an extra term to the series will still have it be convergent.

 

[ardentmed]

OK good point, the function isn't decreasing over that whole integral. really you need to say $\displaystyle \begin{align*} \sum_{n = 3}^{\infty}{ \frac{ \ln{(n)}}{n^2} } \leq \int_2^{\infty} { \frac{\ln{(x)}}{x^2}\,\mathrm{d}x } \end{align*}$ as it is in this region that the entire function is decreasing.

Showing that this integral is convergent (i.e. evaluating the integral) will show that this particular series converges. Adding an extra term to the series will still have it be convergent.

Alright, so 4a must be convergent because the integral -lnx/x - 1/x converges to 1 which is a finite value and thus the series converges.

As for 4d, the integral test proves divergence because ln(x)^2 / 2 = infinity.

On the other hand, you could use L'Hopital's rule and get 0 for 4d which is convergent.

Am I on the right track?

Thanks in advance.

 

[Prove It]

Alright, so 4a must be convergent because the integral -lnx/x - 1/x converges to 1 which is a finite value and thus the series converges.

Well it converges, but not to 1...

As for 4d, the integral test proves divergence because ln(x)^2 / 2 = infinity.

It most certainly does not!

On the other hand, you could use L'Hopital's rule and get 0 for 4d which is convergent.

Which means that the series MAY converge. Again, you need to use a CONVERGENCE TEST. As we have already stated, use the ALTERNATING SERIES TEST!

Where is $\displaystyle \begin{align*} \frac{\ln{(x)}}{x} \end{align*}$ a decreasing function? I sincerely hope you know that a function is decreasing where its derivative is negative...