1
$\begingroup$

I need to show that the following function is continuous over $(0, \infty)$:

$$f(x) = \sum_{n=1}^{\infty}{(\cos(nx))^{n^2}\over{(e^x + x)^n}}$$

I have tried showing that the series is uniformly convergent over said interval, using the Weierstrass M-test:

$${(\cos(nx))^{n^2}\over{(e^x + x)^n}} \le {1\over (e^x + x)^n} $$

And since $x \ge 0$, we have:

$${1\over (e^x + x)^n} \le {1\over (e^x)^n}$$

Now, since $\sum_{n=1}^{\infty}{ {1\over (e^x)^n}} $ is a convergent geometric series, as I understand it, it is enough to claim that $f(x)$ is uniformly convergent, and thus continuous over $(0,\infty)$.

But it seems that I am mistaken. The formal solution I was given included showing that

$${1\over (e^x + x)^n} \le {1\over (e^b)^n + b}$$ in the interval $[b,\infty) $ for every $b \ge 0$.

I would appreciate if someone can explain to me why my solution is wrong, or isn't enough.

Thanks.

$\endgroup$
1
  • 1
    $\begingroup$ Note that the geometrical series you find does not converge for $x=0$ so you can't get uniform convergence on $[0,\infty)$ from this. However to get continuity on $(0,\infty)$ you need to show that it's continuous for any given $x>0$ (but not $0$ itself). Given any $x>0$ there is a $b>0$ such that $b<x$ so you only need to show uniform convergence on $[b,\infty)$ for any $b>0$. $\endgroup$ – Winther Jul 13 '17 at 13:23
1
$\begingroup$

For the Weierstraß -M -test you need an estimation of the form

${1\over (e^x + x)^n} \le a_n$ such that the sequence $(a_n) $ is independent of $x$ !!

(and such that $\sum_{n=1}^{\infty} a_n$ is convergent).

$\endgroup$

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.