Finding limits of summation

unscientific

1. The problem statement, all variables and given/known data

I have attached the question, along with the solution in the picture attached. This is one of the few questions I have encountered that I completely have no idea what the solution is trying to do...

It's like they do not make any sense at all!

Confused by
1. The summation on the right - i have no idea what is the purpose of it

2. The terms inside the brackets are also rather weird in terms of M_k-1 or M_k. The summation on the left - it starts from k = 2, 3, ..... which doesn't even match the terms in the brackets!


2. Relevant equations



3. The attempt at a solution

I tried finding the d' Alembert's ratio but it gave me nonsensical answer of as long as x < 0 the series converges which is not true. If you compare the series to 1/n, the series diverges for x > -1

Will appreciate it if anyone can help.

Curious3141

No attachment.

unscientific

****, thanks for spotting my mistake man. Here's the attachment!

Attached Thumbnails

 

unscientific

help??

SammyS

That solution is rather difficult to understand.

I'll work on understanding it, then attempt to explain it.

Zondrina

Er if you've learned the ratio test, it would help you a lot here. Simply calculate :

lim n->∞ |a_n+1/a_n|

This will tell you whether your series is absolutely convergent, divergent or the test simply fails and you have to employ another test like the integral test or the alternating series test to find your answer. Also if you use this like so :

lim n->∞ |a_n+1/a_n| < 1

You can also determine for what values of x the series converges for as well as things like the radius of convergence.

Hope this helps.

unscientific

bump...help anyone? I'm left with this last question for this chapter!

p = lim ( n → ∞) [ (ln n+1)/(ln n) ]^x < 1

Then i tried to 'ln' both sides, which brings me:

x * ln [ ln(n+1) ] < 0


Which brings me a nonsensical answer of x < 0 which is not true, using a comparison test with 1/n .

Robert1986

I'm guessing you aren't allowed to use the Cauchy Condensation Test, right? Here is a wikipedia link: http://en.wikipedia.org/wiki/Cauchy_condensation_test

Basically, to me, it looks like the answer in the book is "proving" the condensation test for this one special case. I don't think you will get anywhere just doing a straight ratio test.

unscientific

It wasn't even written anywhere in the book! Thanks man, i'll have a go at the link.

SammyS

unscientific,

I don't know if you're still interested, but I do understand the argument given in the link you gave.

First, look at the values of M_k, for several value of k.

k=0: M₀ = 2, since ln(2) ≈ 0.693 and ln(1) = 0 .

k=1: M₁ = 3, since ln(3) ≈ 1.099 and ln(2) ≈ 0.693 .

k=2: M₂ = 8, since ln(8) ≈ 2.079 and ln(7) ≈ 1.946 .

k=3: M₃ = 21 .

k=4: M₄ = 55 .

k=5: M₄ = 149 .
...

Looking at the nested summation:

[itex]\displaystyle \large S_1=\sum_{k=1}^{\infty} \left(\sum_{r_k=1+M_{k-1}}^{M_k}\frac{1}{\left(\ln(M_k)\right)^X} \right)[/itex]

We see that when k = 1, r₁ goes from 3, to 3:

[itex]\displaystyle \frac{1}{\ln(M_1)}=\frac{1}{\ln(3)}<\frac{1}{\ln(n)}\,,[/itex] for n=2 .

When k = 2, r₂ goes from 4, to 8:

[itex]\displaystyle \frac{1}{\ln(M_2)}=\frac{1}{\ln(8)}<\frac{1}{\ln(n)}\,,[/itex] for n=3,...,7 .

When k = 3, r₃ goes from 9, to 21:

[itex]\displaystyle \frac{1}{\ln(M_3)}=\frac{1}{\ln(21)}<\frac{1}{\ln(n)}\,,[/itex] for n=8,...,20 .

etc.


There is an error in the last expression for S₁. M_k - M_k-1 is only approximately equal to (1 - e^-1)M_k, but they are equal in the limit k→∞ . So the final result is valid.

There is a somewhat more straight forward way to achieve the desired result, using a modified version of this method.

unscientific

First of all, thank you for posting your solution. I have read everything and I understand every step, but i am still lost.

I don't see how this answers the question and the purpose of this summation eludes me.. In short, I understand the steps, but I don't see how this is even related to the question at all

The only thing i see is that this summation does the job of summating from k = 1 to k = ∞. But how did they even conceive of this in the first place?!

SammyS

S₁ < S.

The ratio test for S₁ shows that S₁ diverges, for X considered here, i.e. X ≥ 1, which corresponds to x ≤ -1 .

How did they come up with this? I don't know off hand, but it's basically regrouping the
terms of the sum. A similar, but simpler, process can be used to show that [itex]\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}[/itex] diverges.

unscientific

Thank you, I now see how this solves to the question! This is one of the times when i look at the solution and get intrigued by how the person even came up with this..

unscientific

I figured this must be the thought process of the author:

1. He first had the idea in mind to find a series S₁ < S

2. He saw that every term in the series gets smaller and smaller and sought to find a series that used the grouping method.

2. Next he fiddled here and there and finally came up with the governing equation: ln(M_k) > k

3. Then he tried evaluating certain values of k and tried to match the number of terms for each value of k, like for example behind ln 8 there's ln 3, ln 4, ..... ln 7. So the summation would be from M_k-1 + 1 to M_k.

4. Then by ratio test, Voilà!


What puzzles me still, is how he came up with ln M_k > k ..

SammyS

It looks to me as if the person was relating the S sum to the sum [itex]\displaystyle \sum\frac{1}{n}\,,[/itex] so he based S₁ on Ʃ(1/n).

Looking at 1/ln(M_k) for several values of k appears to confirm this.

k=1: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(3)}\approx\frac{1}{1.099}<\frac{1}{1}=\ frac{1}{k}[/itex]

k=2: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(8)}\approx\frac{1}{2.079}<\frac{1}{2}=\ frac{1}{k}[/itex]

k=3: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(21)}\approx\frac{1}{3.045}<\frac{1}{3}= \frac{1}{k}[/itex]

k=4: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(55)}\approx\frac{1}{4.007}<\frac{1}{4}= \frac{1}{k}[/itex]

k=5: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(149)}\approx\frac{1}{5.003}<\frac{1}{5} =\frac{1}{k}[/itex]

k=6: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(404)}\approx\frac{1}{6.0014}<\frac{1}{6 }=\frac{1}{k}[/itex]

k=7: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(1097)}\approx\frac{1}{7.0003}<\frac{1}{ 7}=\frac{1}{k}[/itex]

k=8: [itex]\displaystyle \frac{1}{\ln(M_k)}=\frac{1}{\ln(2981)}\approx\frac{1}{8.00001}<\frac{1} {8}=\frac{1}{k}[/itex]

As you can see, as k increases, ln(M_k) tends to get closer and closer to integer values.

unscientific

I see, thanks!! I will try to think harder next time :)

SammyS

I doubt that this was a very easy solution to come up with.

It certainly isn't all that easy to understand --- not very transparent.

I'll try to get around to showing what I think is a somewhat clearer soliution in the next day or two.