Convergence of the Sum of Tan(1/n)/(1+n) for n=1 to Infinity?

[kskiraly]

Homework Statement

sum of tan(1/n)/(1+n) for n=1 to infinity

Homework Equations

The Attempt at a Solution

I tried using the ratio test and the comparison/limit comparison tests but can't think on anything to compare it to.

 

[mybsaccownt]

I'm far from a math wiz, but here's how I see it:tan(x) will always be between -1 and 1,

but tan (1/n) will always be positive so it will be between 0 and 1

(well, actually for this problem, the max value of the numerator is tan(1) and it continues to decrease)

and the denominator will be at the very least 2 and it will continue to increase

so your fraction will continue to get smaller and smaller

in fact, it will always be smaller than 1/n

since the top will be at most 1 and the the denominator will be greater than n

so, tan(1/n)/(n+1) < 1/nand you know that 1/n converges...so by comparison, tan(1/n)/(1+n) must also converge

 

[d_leet]

I'm far from a math wiz, but here's how I see it:


tan(x) will always be between -1 and 1,

Are you sure about that? What does tan(x) approach as x approaches pi/2 from the left?

 

[d_leet]

and you know that 1/n converges...so by comparison, tan(1/n)/(1+n) must also converge

The SEQUENCE with general term 1/n converges, does the series?

 

[gammamcc]

Well.. sin(1/n) is approximately 1/n for large n. I would do some sort of error estimate for tan(1/n) = 1/n + error to see if the the series CONVERGES absolutely. (Error estimates for sin(1/n) may suffice.)

Watch out for comparing to the famously DIVERGING harmonic series.

 

[mybsaccownt]

d_leet, ok tan(x) does approach infinity when cos(x) approaches 0

and

"The harmonic series diverges, albeit slowly, to infinity"

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

well...ok, I forgot about the n^r r > 1 condition for convergence

but that's why I put the disclaimer about not being a math wiz but...BUT...the series in question does converge

<---has super maple skills

 

[kskiraly]

Thanks so much but doesn't 1/n diverge by p-series because p=1. remember p-series diverges where 1/(n^p) p<=1

 

[kskiraly]

mybsaccownt sorry about the last post i didn't fully read your last one.