# Thread: Some problems with series

1. ## Some problems with series (Still need help)

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.

2. 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

3. 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.

4. 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.

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

6. 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.

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

8. 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.

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

sorry for mistake, still it gives 0 to me

10. $
\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.

11. $
\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$

12. 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.

13. 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.

14. 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.

15. Thanks! That helped alot.

Page 1 of 2 12 Last
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}