Absolute convergence of a real series

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?

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

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 $N$.

• Nice answer, thank you. – Mel May 25 '13 at 21:35

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 $[-M,M]$.

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