# proving continuity claims

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?

thank you very much for helping

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

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.

• 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. – Jason Jan 14 '18 at 22:59
• @Jason I think you were wrong in your comment. Read carefully my answer: the limit function is continuous at $\;x=1\;$ . – DonAntonio Jan 14 '18 at 23:00
• 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 $1$. – Jason Jan 14 '18 at 23:08
• @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,. – DonAntonio Jan 14 '18 at 23:13

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.