1
$\begingroup$

I need to show that the following series:

$\sum_{i=1}^n (-1)^n\dfrac{x^2+n}{n^2} $

Is uniformly convergent on any bounded interval, but not absolutely convergent for any real $x$. My first thought was to use the Weierstrass M-Test, however this is pointless as if the above series "passed" the test, it would have to be absolutely convergent.

In assessing uniform convergence, I considered partial sums. Is it possible to show that the partial sums can get arbitrarily close together, (in a Cauchy sense) which would then imply the series converges uniformly?

$\endgroup$
  • $\begingroup$ Think about using the Weierstrauss $M$-Test for a bounded interval. Apparently you cannot find a series to dominate the given series for all $x$. $\endgroup$ – James S. Cook May 25 '13 at 21:04
  • $\begingroup$ @ThomasAndrews You're right, sorry. I'll change the title. $\endgroup$ – Mel May 25 '13 at 21:07
2
$\begingroup$

Note that in absolute value, we have, since all but $(-1)^n$ is positive:

$$\sum_{n=1}^N\frac{x^2+n}{n^2}$$

But this is

$$\sum_{n=1}^N\frac{x^2}{n^2}+\sum_{n=1}^N\frac{1}{n}$$

The first term converges, but the second term doesn't. Note this tells the $M$ test is not applicable.

On the other hand, your sum is $${f_N}\left( x \right) = \sum\limits_{n = 1}^N {{{( - 1)}^n}} \frac{{{x^2} + n}}{{{n^2}}} = \sum\limits_{n = 1}^N {{{( - 1)}^n}} \frac{{{x^2}}}{{{n^2}}} + \sum\limits_{n= 1}^N {{{( - 1)}^n}} \frac{1}{n}$$

This converges pointwisely to

$$\mathop {\lim }\limits_{N \to \infty } {f_N}\left( x \right) = \sum\limits_{n = 1}^\infty {{{( - 1)}^n}} \frac{{{x^2}}}{{{n^2}}} + \sum\limits_{n = 1}^\infty {{{( - 1)}^n}} \frac{1}{n} = - \frac{{{x^2}{\pi ^2}}}{{12}} + \log 2$$ Now, look at the difference; we have $$\displaylines{ \left| {f\left( x \right) - {f_N}\left( x \right)} \right| = \left| {\sum\limits_{n = N + 1}^\infty {{{( - 1)}^n}} \frac{{{x^2}}}{{{n^2}}} + \sum\limits_{n = N + 1}^\infty {{{( - 1)}^n}} \frac{1}{n}} \right| \cr \leqslant \left| {\sum\limits_{n = N + 1}^\infty {{{( - 1)}^n}} \frac{{{x^2}}}{{{n^2}}}} \right| + \left| {\sum\limits_{n = N + 1}^\infty {{{( - 1)}^n}} \frac{1}{n}} \right| \cr \leqslant {x^2}\left| {\sum\limits_{n = N + 1}^\infty {\frac{1}{{{n^2}}}} } \right| + \left| {\sum\limits_{n = N + 1}^\infty {{{( - 1)}^n}} \frac{1}{n}} \right| \cr \leqslant M\varepsilon + \varepsilon = \left( {1 + M} \right)\varepsilon \cr} $$

where $M$ is a bound for $x^2$ on any bounded interval, for sufficiently large $N$.

$\endgroup$
  • $\begingroup$ Nice answer, thank you. $\endgroup$ – Mel May 25 '13 at 21:35
2
$\begingroup$

By the limit comparison test, the series does not converge absolutely at any point because the harmonic series diverges and for all $x$, $\lim\limits_{n\to\infty}\dfrac{\dfrac{x^2+n}{n^2}}{\dfrac{1}{n}}=1$.

The series converges uniformly on bounded intervals because $\sum\limits_{n=1}^\infty\dfrac{(-1)^n}{n}$ converges everywhere uniformly (there's no $x$ in it), whereas on the interval $[-M,M]$, the Weierstrass $M$ test can be applied with $\dfrac{x^2}{n^2}\leq \dfrac{M^2}{n^2}$, showing that $\sum\limits_{n=1}^\infty\dfrac{(-1)^nx^2}{n^2}$ converges absolutely and uniformly on $[-M,M]$.

$\endgroup$
1
$\begingroup$

Hint: For uniform convergence, look at the natural estimate of the remainder for series satisfying Leibniz' test (a.k.a the alternating series test).

$\endgroup$

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.