<s>[INST] Proving bounds of a harmonic series Let $p>1$. Prove that the series:
$\sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^p}$
is between $\frac {1}{2}$ and 1.
Any help is appreciated.
Just a challenge problem I was presented and curious on the solution
Thanks
 [/INST] Since $\sum_{n = 1}^\infty (-1)^{n+1}/n^p$ is absolutely convergent for $p > 1$, we may rearrange the terms without affecting the sum. Now
$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^p} = \left(1 - \frac{1}{2^p}\right) + \left(\frac{1}{3^p} - \frac{1}{4^p}\right) + \cdots$$
and each term in parentheses is positive, so the sum of the series is greater than $1 - 1/2^p$, which is greater than $1/2$. Since
$$\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^p} = 1 - \left(\frac{1}{2^p} - \frac{1}{3^p}\right) - \left(\frac{1}{4^p} - \frac{1}{5^p}\right) - \cdots$$
and each term in parentheses is positive, it follows that the sum of the series does not exceed $1$. So $\sum_{n = 1}^\infty (-1)^{n+1}/n^p$ lies between $1/2$ and $1$.
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