Re: question regarding convergence of series


> alon78 wrote:
> > I need help showing whether or not the series
> sigma[(cosn)/(cosn+sqrtn)] converges.
> > Thanks in advance for any assistance.
> >
> > -Alon
>
"johnsoi" <johnsoi@xxxxxxxxxxxx> wrote in message
news:1115908842.590086.72000@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Your Series diverages. here is how it is done.
>
> We Will use the Comparsion Test ,Limit Comparsion Test, and a p-series
> to find the answer. I am assuming those test are understood by reader
> if not I can explain.
>
> Look at the Numerator of you infinite sum.
>
> Cos(x)
>
> For any x Cos(x) will always be greater than -1 and less than 1
> -1 <= Cos(x) <= 1
> (<= means less than or equal to)
>
> Divide Each term by [Cos(x) + Sqrt(n)]
>
> -1/[Cos(x)+ Sqrt(n)] <= Cos(x)/[Cos(x)+ Sqrt(n)]
>
> So if -1/[Cos(x)+ Sqrt(n)] diverages, then something great than a
> diveragent series also diverages (Comparsion test).
>
> -1/[Cos(x)+ Sqrt(n)] ~ 1/Sqrt(n) For Large n (~ approx. equal to)
>
> Take the limit( {-1/[Cos(x)+ Sqrt(n)]} / {1/Sqrt(n)} ) as n->inf
>
> Which is equal to one, Thus -1/[Cos(x)+ Sqrt(n)] behaves like 1/Sqrt(n)
>
> 1/Sqrt(n) is a p-series with p=1/2 which is < 1 so it will diverage.
>
> If 1/Sqrt(n) diverages then -1/[Cos(x)+ Sqrt(n)]
>
> And since -1/[Cos(x)+ Sqrt(n)] is divergent and Cos(x)/[Cos(x)+
> Sqrt(n)] is greater than a divergent series it also diverges.
>
> I hope that helps.

No, that won't help. You can use the Comparison Test only on series with
nonnegative terms. The series
sum( cos(n)/(cos(n) + sqrt(n)), n = 1 to infinity)
behaves more like an alternating series, so I suspect that it converges.
However, I will have to think about this a while to come up with a proof.

________________________________
Eric J. Wingler (wingler@xxxxxxxxxxxx)
Dept. of Mathematics and Statistics
Youngstown State University
One University Plaza
Youngstown, OH 44555-0001
330-941-1817



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Relevant Pages

  • Re: question regarding convergence of series
    ... We Will use the Comparsion Test,Limit Comparsion Test, and a p-series ... diveragent series also diverages. ... Sqrt] is greater than a divergent series it also diverges. ...
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  • Re: question regarding convergence of series
    ... > We Will use the Comparsion Test,Limit Comparsion Test, and a p-series ... > diveragent series also diverages. ... > Sqrt] is greater than a divergent series it also diverges. ...
    (sci.math)