Some problems with series

**Hweengee** · November 12th 2008, 02:37 AM

How do I determine the convergence/divergence of the following:

$ \sum _{n=1}^{\infty }\left({\sqrt[n] 3}-1 \right) $

I can't use limit comparison test with ${\sqrt[n] 3}$ because it gives me 0. Root and ratio tests don't seem appropriate and I'm struggling to find a divergent series which is smaller than this to compare it with. Would it be possible to compare it with $\sum _{n=1}^{\infty }\left(\ln\left(n \right)-1 \right)$ ?

As for

$\sum _{n=3}^{\infty }\frac{1}{\left( lnln\left(n \right) \right)^{\ln \left( n \right) }}$

I'm more or less completely stuck.

Thanks for the help in advance.

**Mathstud28** · November 12th 2008, 03:00 AM

Originally Posted by Hweengee

How do I determine the convergence/divergence of the following:

$ \sum _{n=1}^{\infty }\left({\sqrt[n] 3}-1 \right) $

I can't use limit comparison test with ${\sqrt[n] 3}$ because it gives me 0. Root and ratio tests don't seem appropriate and I'm struggling to find a divergent series which is smaller than this to compare it with.

As for

$\sum _{n=3}^{\infty }\frac{1}{\left( lnln\left(n \right) \right)^{\ln \left( n \right) }}$

I'm more or less completely stuck.

Thanks for the help in advance.

For the first one, consider what power series $\sqrt[n]{3}$ equals.

For the second one

Consider

$\frac{1}{\ln(\ln(n))^{\ln(n)}\leq{\frac{1}{\ln\lef t(\frac{1}{n}\right)^{\ln(n)}$

That is a convergent alternating series

**toraj58** · November 12th 2008, 03:06 AM

i use Root Test to solve this problem:

$\lim_{n \to \infty} \sqrt[n]{\sqrt[n] {3} - 1}$

$= \lim_{n \to \infty} \sqrt[n] {\sqrt[n] {3} (1- \frac {1}{\sqrt[n] {3}})}$

$= \lim_{n \to \infty} {\sqrt 3}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}$ = 0

because 0 < 1 then the mention series Converge.

**Mathstud28** · November 12th 2008, 03:20 AM

Originally Posted by toraj58

i use Root Test to solve this problem:

$\lim_{n \to \infty} \sqrt[n]{\sqrt[n] {3} - 1}$

$= \lim_{n \to \infty} \sqrt[n] {\sqrt[n] {3} (1- \frac {1}{\sqrt[n] {3}})}$

$= \lim_{n \to \infty} {\sqrt 3}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}$ = 0

because 0 < 1 then the mention series Converge.

You performed your limit wrong, the root test gives one, which is indeterminate.

**toraj58** · November 12th 2008, 03:40 AM

are you sure it gives 1, why?....i got 0!

**Hweengee** · November 12th 2008, 03:47 AM

yea root test gives a value of one. I'm not really sure what the power series representation of $\sqrt [n] 3$ is, but I'll go look it up. Thanks.

**toraj58** · November 12th 2008, 03:52 AM

i don't understand how both of you got 1, please explain!

**Hweengee** · November 12th 2008, 04:07 AM

Originally Posted by toraj58

i use Root Test to solve this problem:

$\lim_{n \to \infty} \sqrt[n]{\sqrt[n] {3} - 1}$

$= \lim_{n \to \infty} \sqrt[n] {\sqrt[n] {3} (1- \frac {1}{\sqrt[n] {3}})}$

$= \lim_{n \to \infty} {\sqrt 3}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}$ = 0

because 0 < 1 then the mention series Converge.

the 2nd and 3rd step aren't equivalent.

**toraj58** · November 12th 2008, 04:30 AM

$= \lim_{n \to \infty} {3^{\frac {1}{n^2}}}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}$ = 0

sorry for mistake, still it gives 0 to me

**Hweengee** · November 12th 2008, 05:13 AM

$ \lim_{n \to \infty} {3^{\frac {1}{n^2}}}\sqrt[n] {1- \frac {1}{\sqrt[n] {3}}} $

$\lim_{n \to \infty} {3^{\frac {1}{n^2}}}=1$

$\lim_{n \to \infty} \sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}=1$

So their product must also be 1. I don't get how you get 0.

**toraj58** · November 12th 2008, 05:56 AM

$ \lim_{n\to\infty}{\frac {1}{\sqrt[n] 3}} = \lim_{n\to\infty} \frac {1}{3^{\frac {1}{n}}} = \frac {1}{3^0} = \frac {1}{1} = 1 $

so

$ \lim_{n \to \infty} \sqrt[n] {1- \frac {1}{\sqrt[n] {3}}}= \lim_{n \to \infty} \sqrt[n] {1 - 1} = 0$

**Hweengee** · November 12th 2008, 08:06 AM

yes, but ${\frac {1}{n}}$ tends to 0 as n tends to infinity. so you have an indeterminate form of $0^0$ which is in this case 1. anyway i believe this series is divergent.

**Laurent** · November 12th 2008, 08:58 AM

You can write $\sqrt[n]{3}-1=e^{\frac{\ln 3}{n}}-1\sim_{n\to\infty} \frac{\ln 3}{n}$
(Because $\frac{e^x-1}{x}\to_{x\to 0} \exp'(0)=1$ or just (this is the same) because of the usual Taylor expansion of the exponential at zero: $e^x=1+x+o(x)$ )
As a consequence, since $\sum_n \frac{1}{n}$ diverges, and we deal with series with positive coefficients, you can conclude that your series diverges as well.

**Laurent** · November 12th 2008, 09:11 AM

I have another elementary proof:

Using the identity $a^n-1=(a-1)(1+a+\cdots+a^{n-1})$ , we have:
$2=3-1=\left(\sqrt[n]{3}\right)^n-1=(\sqrt[n]{3}-1)\left(1+\sqrt[n]{3}+\left(\sqrt[n]{3}\right)^2 +\cdots+\left(\sqrt[n]{3}\right)^{n-1}\right)$ $\leq (\sqrt[n]{3}-1) 3n$ (since every of the $n$ terms in the sum is less than 3), so that $\sqrt[n]{3}-1\geq\frac{2}{3n}$ . However, $\sum_n\frac{1}{n}$ diverges.

**Hweengee** · November 12th 2008, 09:13 AM

Thanks! That helped alot.

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