Series

**Mathstud28** · May 15th 2008, 06:39 AM

I was practicing up on my convergence tests using one of those archaic Dover books that are chalk full of good info but are near unreadable and formal/difficult to boot. And I was doing the "Very difficult" convergence excersises and I came up with two questions

The first is that this was supposed to be one of the hardest problems in the book...so I must be doing something wrong..Don't get me wrong..I have condfidence in myself and my answer but this can't be this easy if this is one of the hardest in the book

"Find the convergence for $\sum_{n=0}^{\infty}\bigg[\sqrt[n]{n}-1\bigg]^n$ "

Applying the root D'alembert's root test we get

$\lim_{n\to\infty}\bigg[\sqrt[n]{n}-1\bigg]^{\frac{n}{n}}=\lim_{n\to\infty}\sqrt[n]{n}-1=0$

$\therefore$ series is convergent? Is there soemthing wrong witht that?

Secondly

I dont know what to call this...is this actually a formal definition...I am about 99% sure its correct

Let $\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}b_n\pm{c _n}$

If either $\sum_{n=0}^{\infty}b_n$ or $\sum_{n=0}^{\infty}c_n$ diverges we may say that $\sum_{n=0}^{\infty}a_n$ diverges

Is it sufficent to say that as a justification?

**colby2152** · May 15th 2008, 07:04 AM

$\lim_{n \rightarrow \infty} n^{1/n} = 1$ , so the convergence to zero is correct.

I do not have a book in front, but the logic behind adding a convergent and divergent sequence/series is fairly simple, and the result is a divergent sequence/series.

**Moo** · May 15th 2008, 07:58 AM

Hello !

Originally Posted by colby2152

$\lim_{n \rightarrow \infty} n^{1/n} = 1$ , so the convergence to zero is correct.

I do not have a book in front, but the logic behind adding a convergent and divergent sequence/series is fairly simple, and the result is a divergent sequence/series.

Adding a divergence series doesn't always yield a divergence series :-)

$\frac 1n$ and $-\frac 1n$

Applying the root D'alembert's root test we get

This is Cauchy's one, not d'Alembert's

**ThePerfectHacker** · May 15th 2008, 10:39 AM

Originally Posted by Mathstud28

"Find the convergence for $\sum_{n=0}^{\infty}\bigg[\sqrt[n]{n}-1\bigg]^n$ "

You do not need the ratio test here. Just the fact that $n^{1/n}\leq 2$ for $n\geq 2$ . This means $1\geq \tfrac{1}{2}n^{1/n}$ .

Now, $0\leq \left( n^{1/n} - 1\right)^{n} \leq \left( n^{1/n} - \tfrac{1}{2}n^{1/n} \right)^n = n\left( \frac{1}{2} \right)^n$ .

And $\sum_{n=0}^{\infty} n\left( \frac{1}{2}\right)^n$ certainly converges.

**Mathstud28** · May 15th 2008, 11:49 AM

Originally Posted by Moo

Hello !

Adding a divergence series doesn't always yield a divergence series :-)

$\frac 1n$ and $-\frac 1n$

This is Cauchy's one, not d'Alembert's

Haha..I never learned the names of the inventors of the tests...I just tried to remember...your right its D'alembert's Ratio test isnt it?

And I did omit something....I had it originally in the post but accidentally erased it...it should have read
""Secondly

I dont know what to call this...is this actually a formal definition...I am about 99% sure its correct

Let

If either

or

diverges we may say that

diverges

Is it sufficent to say that as a justification?NOTE when I say or I mean or...not and/or...""

So since your example is two divergent series that is not what I am talking about

I am talking about something akin to this

$\sum_{n=1}^{\infty}\frac{n+\sqrt{n}}{n}=\sum_{n=1} ^{\infty}\frac{1}{n}+\sum_{n=1}^{\infty}\frac{1}{n ^{\frac{3}{4}}}$

I know that this concept is correct but I just want to make sure there arent any technical/formal errors in it...for the justification as to why this is divergent I would state

"Since $\frac{1}{n}$ is divergent and $\frac{1}{n^{\frac{3}{4}}}$ is convergent..their sum is divergent"

I know that is right but would you think a teacher would accept it..as well as more importantly is it clear to a student if it was taught?

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