Convergence or Divergence? Tips for Solving Tricky Series Problems

[weesiang_loke]

Homework Statement

inf
$$\sum$$ ( $$\sqrt[n]{n}$$ - cos ($$\frac{1}{n}$$ ) *** edited
n=1

the series converges or diverges?

Homework Equations

ratio test, dirichlet's test, comparison test, etc

The Attempt at a Solution

i tried a lot of method but still cannot get the answer...
from what i know, limit n to inf of n^(1/n) is 1 but limit n to inf of cos (1/n) is also one. so the divergence test cannot applied here.
besides, i tried the root test and ratio test but still cannot get the answer?
any help or hint would be appreciated... thanks

 

[micromass]

What is

$$\lim_{n\rightarrow +\infty}{\sqrt[n](n)+\cos(1/n)}$$??

 

[weesiang_loke]

not limit but summation, a series

 

[weesiang_loke]

even if you take limit that way the answer will be zero. so it is not conclusive by divergence test..

 

[JG89]

The limit of n^{1/n} + cos(1/n) is NOT zero

 

[weesiang_loke]

limit n to inf n^(1/n) is one, right?
and limit n to inf cos(1/n) = cos (0) = 1.
so the sum of two limit is zero. pls tell me what i did wrong.. thanks

 

[Gib Z]

1+1 = 2, not 0.

 

[weesiang_loke]

sorry i copied the wrong question. it should be minus instead of plus

 

[micromass]

What about the integral test? I believe that may be conclusive...
Or the Cauchy condensation test?

 

[weesiang_loke]

for integral test... we need to at least to integrate (x^(1/x) - cos (1/x)). but i can't find a way to integrate x^(1/x) or cos (1/x)...
(may be my current understanding of calculus is not sufficient for the integration)

and i am not sure how to apply the Cauchy condensation test...
can you pls show me how to begin...
thank you very much...

 

[micromass]

Hmm, you are correct in that the integral test would be rather messy :grumpy: to bad...

But I still believe that the Cauchy condensation test would work. It says that

$$\sum{u_n}~\text{converges iff}~\sum{2^nu_{2^n}}~\text{converges}$$

So, your series converges iff

$$\sum{2^n(\sqrt[n]{2^n}-\cos(2^{-n}))}$$

converges. I believe that this simplifies the integral a bit...

 

[Gib Z]

This question is far too difficult for your level. The integral test or Cauchy's condensation tests won't work here, its not even immediately clear if the terms are monotonically decreasing, which are requirements for those tests. There is no simple solution, but here's a sketch of the solution for other peoples benefit.

$$\sqrt[n]{n} = e^{ \frac{\log n}{n}} = 1 + \frac{ \log n}{n} + O\left( \frac{\log^2 n}{n^2} \right)$$

and $$\cos \left( \frac{1}{n} \right) = 1 + O\left(\frac{1}{n^2}\right)$$

so $$\sqrt[n]{n} - \cos \left( \frac{1}{n} \right) = \frac{\log n}{n} + O\left( \frac{\log^2 n}{n^2} \right)$$, and since $\sum \frac{\log n}{n}$ diverges (comparison with harmonic series, Cauchy Condensation test or integral test all work easily for that), the sum in question diverges as well.

 

[micromass]

But, Cauchy's condensation test DOES work here! It gives us that we need to check convergence of

$$\sum{2^{n+1}-2^n\cos(1/2^n)}$$

Since $$2^n\cos(1/2^n)\leq 2^n$$, we have that

$$\sum{2^{n+1}-2^n}\leq \sum{2^{n+1}-2^n\cos(1/2^n)}$$

and the left sum diverges, so the right sum does to...

 

[weesiang_loke]

I think i got it now.
Thanks for the help, micromass, Gib Z and JG89.
Both methods actually work.

God bless you all and have a nice day.

Cheers,
weesiang_loke