1
$\begingroup$

i am having trouble with two claims i don't know how to prove or disprove:

a)$f(x)=\sum_{n=1}^\infty \arctan\left(\frac{2x}{x^2+n^3}\right)$ is continuous in R.

b)$\sum _{n=1}^\infty (1-x)x^n$ is uniformly converge in [0,1]

what i tired:

a)i don't know how to prove it. how do i prove that this sequence is continuous in R? should i use the taylor approximation of it or it consists of 2 continuous functions in R and then deduct it's continuous in R? knowing that the taylor approximation of arctanx is $\sum_{n=1} ^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$ is continuous in R and then show that $\sum_{n=1} ^\infty\frac{2x}{x^2+n^3}$ is contiuous in r? and then assert it's continuous in R because it's a composite function of 2 functions that are continuous in R? is there any smart and elegant way to do it otherwise?( if it was a finite set i would've tried to test if $\sum_{n=1} ^\infty(\int_a^bu_n(x)dx)=\int_a^b(\sum_{n=1} ^\infty u_n(x))dx$ (putting the domain and sequence inside)

b)(using abel's test or weirstrauss uniform convergence test would make it easier, however i am wonderig how to use cuchy's test to prove it). for the sum of sequences to uniformly converge it has to follow: $|\sum_{k=n+1}^mu_k(x)| < \epsilon$,so $|\sum_{k=n+1}^m x^n(1-x)|< \epsilon$(can i use $x^n$ to assert that $x^n <\epsilon$? how to formally prove it?

please help me if you can. having a hard time with it.

thank you very much for helping

$\endgroup$
  • 1
    $\begingroup$ Hint for (b): if the convergence was uniform, the limit function would be continuous; prove this is not the case at $x=1$. $\endgroup$ – Jason Jan 14 '18 at 22:46
1
$\begingroup$

Observe that for any $\;x,y\in\Bbb R\;$ , we have that $\;|\arctan x-\arctan y|\le|x-y|\;$ (Use the mean value theorem...), and thus taking $\;y=0\;$ we get that $\;\arctan x|\le |x|\;,\;\;x\in\Bbb R\;$ , so

$$\left|\arctan\frac{2x}{x^2+n^3}\right|\le\left|\frac{2x}{x^2+n^3}\right|,\,\,\,\text{and since}\;\;\sum_{n=1}^\infty\frac{2x}{x^2+n^3}$$

converges absolutely for all $\;x\in \Bbb R\;$ , you get what you want with Weierstrass M- Theorem.

For the other one: for $\;x=0\;$ the series is zero, and for $\;x\in [0,1)\;$ we get

$$\sum_{n=1}^\infty(1-x)x^n=\sum_{n=1}^\infty\left(x^n-x^{n+1}\right)=\frac x{1-x}-\frac{x^2}{1-x}=x$$

so the limit function is

$$f(x)=\begin{cases}x,&x\in[0,1)\\{}\\0,&x=1\end{cases}\;\;\implies\;\text{the limit isn't continuous and thus the convergence is}$$

only pointwise but not uniform.

$\endgroup$
  • 1
    $\begingroup$ Absolute convergence for each $x$ is not sufficient to apply the Weierstrass M-test. Indeed, for (b) we do not have uniform convergence, as I explained in a comment. $\endgroup$ – Jason Jan 14 '18 at 22:59
  • $\begingroup$ @Jason I think you were wrong in your comment. Read carefully my answer: the limit function is continuous at $\;x=1\;$ . $\endgroup$ – DonAntonio Jan 14 '18 at 23:00
  • $\begingroup$ Start from $n=0$, then. This doesn't affect whether convergence is uniform, and in this case the function value at $1$ is zero while the limit is $1$. $\endgroup$ – Jason Jan 14 '18 at 23:08
  • $\begingroup$ @Jason Why would we begin with $\;n=0\;$ when it is written that the sum begins with $\;n=1\;$ ?? But you are right, I was wrong: the limit of that sum for any $\;x\neq1\;$ is simply $\;x\;$ , so the function's simply $\;f(x)=x\;,\;\;x\in [0,1)\;,\;\;f(1)=0\;$ , discontinuous at $\;x=1\;$ ...Thanks, aI shall edit,. $\endgroup$ – DonAntonio Jan 14 '18 at 23:13
1
$\begingroup$

For (a), separate into cases of $x_0=0$ and $x_0\neq0$. For the first case, from $|\arctan(x)|\le|x|$, one finds $$|f(x)|\le\sum_{n=1}^\infty\left|\frac{2x}{x^2+n^3}\right|=|x|\cdot\sum_{n=1}^\infty\frac2{x^2+n^3}\to0$$ as $x\to0$. Since $f(0)=0$, this shows $f$ is continuous at $0$. For $x_0\neq0$, fix $0<\epsilon<|x_0|$; for $x\in[x_0-\epsilon,x_0+\epsilon]$ , one has $$\left|\frac{2x}{x^2+n^3}\right|\le\frac{2(|x_0|+\epsilon)}{n^3}$$ which is summable and independent of $x$. Thus, using again $|\arctan(x)|\le|x|$, we deduce that the series $$\sum_{n=1}^\infty\arctan\left(\frac{2x}{x^2+n^3}\right)$$ converges absolutely and uniformly on $[x_0-\epsilon,x_0+\epsilon]$ by the Weierstrass M-test, and in particular, $f$ is continuous at $x_0$.

For (b), one should be suspicious of the continuity since $\sum_nx^n$ blows up as $x\to1$. Indeed, for $x\in[0,1)$, one can remove $(1-x)$ as a common factor and find $$\sum_{n=1}^\infty(1-x)x^n=(1-x)\sum_{n=1}^\infty x^n=(1-x)\cdot\frac{x}{1-x}=x,$$ and yet clearly the series sums to zero for $x=1$. If the convergence were uniform, the limit function would be continuous; since this is not the case, we conclude that the convergence is not uniform.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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