I am to show that the following series converges pointwise, then uniformly. I am aware that uniform convergence implies pointwise convergence, but I have to show the pointwise convergence first and separately.

$\sum_{n=1}^{\infty}log(1+\frac{x}{n^3}), \hspace{2cm} -1 < x <1$

I have started out by looking at the case of $x=0$, in which case the sum obviously converges to zero.

I am however, not entirely sure how to approach the series, as I can't seem to find another convergent series to use the comparison test on. I have tried doing a ratio test, but it ends up inconclusive, since


As for the uniform convergence, I suppose I have to find a convergent majorantseries. But seeing as I struggle to determine a convergent series for a comparison test relating to the pointwise convergence, I am also at loss here.

Any input is appreciated.

  • $\begingroup$ Hint: For $x>0$, the Taylor series for $\log(1+z)$ implies that $\log(1+\frac{x}{n^3})<\frac{x}{n^3}$. $\endgroup$ – Michael Burr Jun 4 '18 at 11:57
  • $\begingroup$ Sorry. My previous comment was gibberish. I have a hard time seeing why it is implied from the taylor series for $log(1+x)$. Would you mind elaborating? $\endgroup$ – user7924249 Jun 4 '18 at 13:24

For $0 \leqslant x < 1$ we have

$$\left|\log\left(1 + \frac{x}{n^3}\right)\right| = \log\left(1 + \frac{x}{n^3}\right) = \int_1^{1+x/n^3} \frac{dt}{t} \\\leqslant \frac{1+x/n^3-1}{1} = \frac{x}{n^3} \leqslant \frac{1}{n^3}$$

For $-1 < x < 0$ we have

$$\left|\log\left(1 + \frac{x}{n^3}\right)\right| = -\log\left(1 + \frac{x}{n^3}\right) = \int_{1+x/n^3}^{1} \frac{dt}{t} \\\leqslant \frac{1 - (1+x/n^3)}{1+x/n^3} = \frac{-x/n^3}{1+x/n^3} = \frac{1}{n^3/(-x) - 1},$$

and for all $n >1$ and $-1 < x < 0$,

$$\left|\log\left(1 + \frac{x}{n^3}\right)\right| < \frac{1}{n^3 - 1}$$

Thus, the $n$-th term is bounded above in absolute value by $1/(n^3-1)> 1/n^3$ for all $n> 1$ and for all $x \in (-1,1)$ and $\displaystyle\sum_{n > 1}\frac{1}{n^3-1}$ converges.

By the Weierstrass test the series is uniformly convergent.

  • $\begingroup$ Very interesting. Thank you. What is the argument for the first inequality? The one where you state the integral is less than the length of the interval. $\endgroup$ – user7924249 Jun 4 '18 at 19:56
  • $\begingroup$ The inequality $x/(1+x) < \log(1+x) < x$ for $-1 < x $ is well known and an application here gives you the bound you need to show uniform (and pointwise) convergence via the Weierstrass M-test. I'm simply using the inequality for that purpose and at the same time justifying it with the integral definition of $\log$. You could use some other approach like $e^x > 1+x \implies \log(1+x) < x$ just as well. To bound the integral I'm simply using $a \leqslant t \leqslant b \implies 1/b \leqslant 1/t \leqslant 1/a \implies \int_a^b (1/t)\, dt < (1/a)(b-a)$. $\endgroup$ – RRL Jun 4 '18 at 20:15
  • $\begingroup$ The integral is not simply less than the length of the interval. It is less than the length of the interval times an upper bound for the integrand, which happens to be $1$ for the first case and $1/(1+x/n^3)$ for the second. $\endgroup$ – RRL Jun 4 '18 at 20:17
  • $\begingroup$ That makes perfect sense! Fantastic! $\endgroup$ – user7924249 Jun 4 '18 at 20:22

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