# An estimate for $\ln(1+f(x))$ using Taylor expansion

A crucial skill for every aspiring analyst (like myself) is confidence in estimation - knowing when, where, and how to use tools like Big-and-little-O to gain quick upper bounds. I'm trying to push myself to get better at it, but I'm still a little hesitant sometimes. Here's an example: I need to determine if the following series converges uniformly

$$\sum_{n=1}^\infty f_n(x)=\sum_{n=1}^\infty\frac{\ln\left(1+\frac{\sin^2(nx)}{n^2}\right)}{n}$$

I'm pretty sure it does, and my reasoning is as follows: if $n>1$, $\sin^2(nx)/n^2<1$ for all $x\in\Bbb{R}$, and hence we can use the Taylor series for $\ln(1+z)$ about $z=0$ to obtain the following:

$$\left\vert\ln\left(1+\frac{\sin^2(nx)}{n^2}\right)\right\vert\leq\frac{1}{n^2}+\frac{C}{n^4}\tag{\dagger}$$ where $C$ is a constant independent of $x$. Thus,

$$\left|\,f_n(x)\right|\leq\frac{1}{n^3}+\frac{C}{n^5}$$and since $$\sum_{n=1}^\infty\left( \frac{1}{n^3}+\frac{1}{n^5}\right)<\infty,$$the series converges uniformly.

My question is: is $(\dagger)$ correct, and do I have the constant in the right place (or does it matter)? I used the fact that $\ln(1+z)=z+O(z^2)$ to obtain $(\dagger)$.

-
@msh210 thanks for the copy editing...studying for quals has really cooked my noodle. –  icurays1 Jul 31 '13 at 19:21
You're welcome, but my edit suggestion was accepted after you had made a clashing edit -- or so it seems -- so yours was lost. –  msh210 Jul 31 '13 at 19:24
Oh, I was wondering why a line disappeared... –  icurays1 Jul 31 '13 at 19:25

You can actually use the simpler estimate, that says that $$\log(1+x)\leq x$$

Then $$\frac{1}{n}\log \left( {1 + \frac{{{{\sin }^2}nx}}{{{n^2}}}} \right) \leqslant \frac{{{{\sin }^2}nx}}{{{n^3}}}\leqslant \frac 1 {n^3}$$

One should always aim for the simple first!

ADD As per your estimate. First note that your function is always positive and periodic. Moreover, it is bounded. Since it has multiple roots, we know there is a maximum, and it will be an absolute "periodic" maxima. The derivative is $$\frac{{\sin 2nx}}{{n + {n^{ - 1}}{{\sin }^2}nx}}$$ which has roots at (choose $k=0,1$) so $2nx=0,\pi$. At $\pi/2n$ the maxima is $$\log\left(1+\frac 1 {n^2}\right)\leq\frac 1{n^2}$$

which is summable.

-
Ah! Now that is much simpler. Is my $(\dagger)$ still correct though, just out of curiosity? –  icurays1 Jul 31 '13 at 19:37
Yes, it's correct. But as Peter says, try the simplest thing first, only if that doesn't work try the more complicated ones. –  Daniel Fischer Jul 31 '13 at 19:37
Another skill I need to work on =) Thanks! –  icurays1 Jul 31 '13 at 19:39