Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is my reasoning for whether $$F(x)=\int_{0}^{x}\sum_{0}^{\infty}\frac{\cos (nt)}{2^n} \text{d} t$$ is continuous in $\mathbb{R}$ correct?:


I claim it is continuous in $\mathbb{R}$. $\sum_{0}^{\infty}\frac{\cos(nx)}{2^n}$ is uniformly convergent in $\mathbb{R}$ (and the functions in the series are continuous), therefore $$F(x)=\int_{0}^{x}\sum_{0}^{\infty}\frac{\cos(nt)}{2^n} \text{d} t = \sum_{0}^{\infty} \int_{0}^{x}\frac{\cos(nt)}{2^n} \text{d} t=x+\sum_{1}^{\infty} \frac{\sin(nt)}{2^n\cdot n}\; .$$

$\sum_{1}^{\infty} \frac{\sin(nt)}{2^n\cdot n}$ is uniformly convergent in $\mathbb{R}$, so $F(x)$ is a uniformly convergent series of continuous functions, which means it is continuous.

Have I correctly solved this exercise? Thank you!

share|improve this question
Your argument is indeed correct. I would just add some justification as why the series are actually uniformly continuous. –  azarel Jan 28 '12 at 15:30
Looks good. You could save a step by noting that the integrand is continuous, due to uniform convergence, and then appeal to the fundamental theorem of calculus to argue that $F(x)$ is continuous. –  David Mitra Jan 28 '12 at 15:35
Thanks, both of you! David Mitra, I will take note of that in the future. I guess this question is closed, perhaps one of you should submit your comment as an answer? (I can't submit an answer since I am a guest user) –  ro44 Jan 28 '12 at 15:40
add comment

2 Answers

up vote 5 down vote accepted

It looks good, but some additional justifications should be made. At the start (and for the later series) you should state why the series converges uniformly (by the Weierstrass_M-test, e.g.).

Also, to justify that the sum of the series is integrable and that switching the order of summation and integration is valid, you should state that the terms of $G(x)=\sum\limits_{n=0}^\infty {\cos(nt)\over 2^n}$ are integrable over any interval $[0,x]$.

But, can save a few steps in your argument. Since $\sum\limits_{n=0}^\infty {\cos(nt)\over 2^n}$ converges uniformly to $G(x)$, and since the terms of this series are continuous, $G(x)$ is a continuous function. The Fundamental Theorem of Calculus immediately gives you the continuity of $F(x)=\int_0^x G(t)\,dt$.

share|improve this answer
add comment

You could use the M-test as follows.

$$ \displaystyle \sum\limits_{n = 0}^\infty {\left|\frac{{\cos nt}}{{{2^n}}}\right|} < \sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}}}} < 2$$

and do the same for the $ \sin tn $ series to make your proof complete.

Otherwise, the proof is correct.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.