Question #114836
let {an} be a decreasing sequence of positive terms .prove that the series summation an sin (nX) converges uniformily on R if and only if nan tends to 0 as n tends to infinity
1
Expert's answer
2020-05-11T13:41:41-0400

Let {an}\{a_n\} be the decreasing sequence of positive terms. Suppose n=1ansin(nx)\sum_{n=1}^{\infty} a_n \sin(nx)  converges uniformly and let the first nn partial sum of the given series be Sn(x)=r=1narsin(rx)S_{n}(x)=\sum_{r=1}^{n} a_r \sin(rx) , By the hypothesis Sn(x)S_{n}(x) converges uniformly.Now by Cauchy's criteria of uniform convergence, for every ϵ>0,NN\epsilon >0,\exist N\in \mathbb{N} such that nN,mNn\geq N , m \geq N for sufficiently large m,n    m,n \implies

Sn(x)Sm(x)<ϵ,xR|S_{n}(x)-S_{m}(x)|<\epsilon , \forall x \in \mathbb{R}     \iff r=m+1narsin(rx)<r=m+1narsin(rx)|\sum_{r=m+1}^{n} a_r \sin(rx)|<\sum_{r=m+1}^{n}|a_r \sin(rx)| (since, from general triangle inequality),thus

r=m+1narsin(rx)<r=m+1narsin(rx)=r=m+1narsin(rx)|\sum_{r=m+1}^{n} a_r \sin(rx)|<\sum_{r=m+1}^{n}|a_r \sin(rx)|=\sum_{r=m+1}^{n}|a_r|| \sin(rx)| .We also know that, always sin(rx)<1|\sin(rx)|<1     r=m+1narsin(rx)<r=m+1nar=r=m+1nar\iff \sum_{r=m+1}^{n} |a_r||\sin(rx)|<\sum_{r=m+1}^{n} |a_r|=\sum_{r=m+1}^{n} a_r (since, ar>0,rNa_r >0 ,\forall r \in \mathbb{N} .Hence choose ϵ=r=m+1nar    ϵ=r=m+1nar>an(nm)\epsilon = \sum_{r=m+1}^{n} a_r \iff \epsilon = \sum_{r=m+1}^{n} a_r> a_n (n-m) ({an}\because \{a_n\} is decreasing sequence),thus an0,a_n \rightarrow 0 , as nn \rightarrow \infty (since, ϵ\epsilon is finite positive real number).

Hence, we are done.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
12.05.20, 20:50

Dear Sheela John, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

Sheela John
12.05.20, 10:38

Thanks for your help assignment expert

LATEST TUTORIALS
APPROVED BY CLIENTS