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Find the set of real numbers for which f(x) exists, where:

$ a) f(x) = \sum\limits_{n=1}^\infty \left( x + \frac{1}{n} \right)^n;$

$ b) f(x) = \sum\limits_{n=1}^\infty \frac{x}{(1+x^2)^n}; $

$ c) f(x) = \sum\limits_{n=1}^\infty \frac{x+ n \cdot (-1)^n}{x^2+n^2}. $

In each case study the continuity of $f$.

Using number series criteria we can prove that those sets are:

a) $(-1,1)$;

b) $ \mathbb{R} $;

c) $ \mathbb{R} $.

(I would like a confirmation if these are the correct answers, because I might be wrong.)

For the second part, there is a problem. I don't know how to prove that the given series of functions are uniform convergent (if they are). We know that an uniform convergent series of continuous functions is continuous, so if we prove that the given series of functions are uniform convergent, we have all we need. Can somebody help me, please? Thank you!

npatrat
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1 Answers1

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a) You are correct: if $ |x| < M$ for some M, then for high enough $n$, we have $|x + 1/n| < M$, so $\sum_n |x + 1/n|^n$ is dominated by $\sum_n M^n$. If M < 1, then it must converge uniformly on $[-M, M]$. Clearly the series does not converge when $|x| \ge 1$, as the summand does not go to $0$ as $n \to \infty$ then.

b) Your are correct: This is $x\sum_n b^n$, where $b = {1\over 1 + x^2}$, which is well-known to converge if and only if $-1 < b < 1$, which translates into $|x| > 0$. However if $x = 0$, then each term in the original series is $0$, so it converges there as well. Note that since $\sum_n b^n$ is known, you can figure out a simple form for $f$ that can be used to examine continuity without resort to convergence theorems.

c) You are correct: Break it into two parts: $\sum_n \frac{x}{x^2 + n^2}$ is easily seen to converge by comparison with $\sum_n \frac{1}{n^2}$. For $\sum_n (-1)^n\frac{n}{x^2 + n^2}$, note that the summands are decreasing in magnitude for $n > x$. By the alternating series test, this converges for all $x$.

The first part converges uniformly. But I'm not sure about the 2nd part.

Edit:

Let $f_n(x) = \frac{(-1)^nn}{x^2 + n^2}$. Then $$|f_n'(x)| = \left|\frac{(-1)^{n+1}2nx}{(x^2 + n^2)^2}\right| \le \frac{2|x|}{n^3}$$ So on compact intervals $[-M, M]$, $\sum_n f'_n(x)$ converges uniformly to some function $g$. Further, note that $\sum_n f_n(0) = \sum_n \frac{(-1)^n}n = \ln 2$. Therefore by this theorem $\sum_n f_n(x)$ converges uniformly on $[-M, M]$ to a continuous function. Since $M$ is arbitrary, it converges to a continuous (in fact, differentiable) function on all of $\Bbb R$.

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Paul Sinclair
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  • a) OK. You proved that the given series is u.c. on every $[-M,M] \subset (-1,1)$. But, this doesn't mean that it is u.c. on $(-1,1)$. How can I prove that the given series is u.c. on $(-1,1)$? (If it is!!!) – npatrat Nov 10 '15 at 08:18
  • c) Does anybody know how to prove (if it is possible) that the series $\sum\limits_{n \ge 1} (-1)^n \frac{n}{x^2+n^2}$ is uniform convergent? Thank you! – npatrat Nov 10 '15 at 08:20
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    a) You are trying to prove continuity, not uniform convergence. Continuity occurs at points. Any point in $(-1,1)$ is inside $(-M,M)$ for some $M$, so the series converges uniformly on a neighborhood of the point, and therefore to a continuous function at the point. Thus the series converges to a continuous function everywhere in $(-1,1)$. Uniform convergence rarely ever holds on open sets with a point of non-convergence on the border, but fortunately, it doesn't need to hold on the entire set to prove continuity. – Paul Sinclair Nov 10 '15 at 22:00