Troubleshooting Series: Overcoming a Persistent Problem

[rasi]

i tried... i can't get tackle with this problem no matter how hard i try. please help me.

 

[HallsofIvy]

All homework posts should use the "homework template" and must show what you have tried yourself on the problem. rasi, if you do not show what you have tried within a couple of days, I will delete this thread.

 

[rasi]

All homework posts should use the "homework template" and must show what you have tried yourself on the problem. rasi, if you do not show what you have tried within a couple of days, I will delete this thread.

could you explain using the "homework template". i'am a little while ago member in this community.
this is my trial.

 

[Curious3141]

could you explain using the "homework template". i'am a little while ago member in this community.
this is my trial.

First of all, you have to be very careful splitting up infinite series like that (in fact, it doesn't look like you can split it up in this case).

Second of all, the second sum is NOT e. The limit term is e.

This is a tricky sum, and I'm not sure how to proceed either. But I don't think you're on the right track yet, if that helps.

 

[rasi]

First of all, you have to be very careful splitting up infinite series like that (in fact, it doesn't look like you can split it up in this case).

Second of all, the second sum is NOT e. The limit term is e.

This is a tricky sum, and I'm not sure how to proceed either. But I don't think you're on the right track yet, if that helps.

first of all thank for everything...
i have been awaken after the event. i have been send it.

 

[Curious3141]

I can see (from google) that you probably got this from the book "Problems in Real Analysis" by Radulescu et al. I don't have the book myself, but if you have it, does the book provide a solution later on?

 

[Curious3141]

Having "procured" the book and taken a good look at it, I can confirm that the solution is *not* given for this and other problems meant for self-study. The book is not elementary, and I have seen few books with a steeper learning curve, even if each chapter starts off in a deceptively simple fashion.

 

[rasi]

you right. this book is very difficult. is there any book that you know which is starting elementary? and has much examples with solutions. about sequences, series, limit, differentiable, derivative, integral

 

[Ray Vickson]

i tried... i can't get tackle with this problem no matter how hard i try. please help me.

For small x > 0 we have $x - x^2/2 < \ln(1+x) < x - x^2/2 + x^3/3,$ so $$\exp\left(1 - \frac{1}{2n}\right)< \left(1 + \frac{1}{n}\right)^n < \exp\left(1- \frac{1}{2n} + \frac{1}{3n^2}\right).$$

RGV

 

[Curious3141]

For small x > 0 we have $x - x^2/2 < \ln(1+x) < x - x^2/2 + x^3/3,$ so $$\exp\left(1 - \frac{1}{2n}\right)< \left(1 + \frac{1}{n}\right)^n < \exp\left(1- \frac{1}{2n} + \frac{1}{3n^2}\right).$$

RGV

I didn't check this, but even assuming it's right, does it help to sum the series? Establishing convergence is quite easy, it's the sum that's killing me.

 

[Curious3141]

you right. this book is very difficult. is there any book that you know which is starting elementary? and has much examples with solutions. about sequences, series, limit, differentiable, derivative, integral

Sorry - you'd better get advice from the Math majors lurking here who would be more up to date. I'm more of a hobbyist.

 

[Ray Vickson]

I didn't check this, but even assuming it's right, does it help to sum the series? Establishing convergence is quite easy, it's the sum that's killing me.

The inequalities above help to bound the terms of the original series between terms of two divergent series, so (assuming I have not made any errors) I get that the series is divergent. I am curious to know how you established convergence.

RGV

 

[Curious3141]

The inequalities above help to bound the terms of the original series between terms of two divergent series, so (assuming I have not made any errors) I get that the series is divergent. I am curious to know how you established convergence.

RGV

But the terms are e - (1+1/n)^n. The limit of that is zero. So I'm unsure how your method establishes divergence.

 

[Dick]

What Ray Vickson is trying to show is that difference between e and (1+1/n)^n is greater than a term of the order of 1/n. So the difference will diverge like a harmonic series.

 

[HallsofIvy]

could you explain using the "homework template". i'am a little while ago member in this community.
this is my trial.

When you first click on the "new thread" button in any of the homework forums, you get a "template" with things like "statement of the problem" and "Attempt at a solution". You chose to erase those. Don't do that!

 

[Curious3141]

What Ray Vickson is trying to show is that difference between e and (1+1/n)^n is greater than a term of the order of 1/n. So the difference will diverge like a harmonic series.

Thank you. I think I've got it, but I'd like to pin it down. Please check if this is on the right track:

$$\exp\left(1 - \frac{1}{2n}\right)< \left(1 + \frac{1}{n}\right)^n < \exp\left(1- \frac{1}{2n} + \frac{1}{3n^2}\right).$$

$$-\exp\left(1 - \frac{1}{2n}\right)> -\left(1 + \frac{1}{n}\right)^n > -\exp\left(1- \frac{1}{2n} + \frac{1}{3n^2}\right).$$

$$e -\exp\left(1 - \frac{1}{2n}\right)> e -\left(1 + \frac{1}{n}\right)^n > e -\exp\left(1- \frac{1}{2n} + \frac{1}{3n^2}\right).$$

Consider only the right hand (lower bound):

$$S = Ʃ(e -\left(1 + \frac{1}{n}\right)^n) > Ʃ(e -\exp\left(1- \frac{1}{2n} + \frac{1}{3n^2}\right))$$

where the sum is taken over the positive naturals.

So $$S > e(1 - e^{(-\frac{1}{2n} + \frac{1}{3n^2})})$$

$$= e[1 - e^{(-\frac{1}{2n})}.e^{(\frac{1}{3n^2})}]$$

$$> e(\frac{1}{6n} - \frac{1}{72n^2})$$

(that last step involved a tedious Taylor's series and multiplication to the third order. I am justified in the greater than equality because the sum of the remaining terms is a positive quantity).

$$=\frac{e}{6}H - \frac{e\pi^2}{432}$$

(where H is the harmonic sum (divergent) and the second term is calculated using ζ(2) ).

$$= ∞$$

Hence the sum diverges.

Sorry, I realize you and Ray would probably consider this unnecessarily tedious, but I really needed to "see" it. Hope there are no errors in my working.

Thanks, Dick and Ray.