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
 A: 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 $1$.
A: In general, consider the sequence $(a_n)$ which satisfies $$a_n\ge a_{n+1}\ge0,\forall n\in\mathbb{N} \,\,\,\,\,\,\text{and} \,\,\,\,\,\,\lim_{n\to\infty}a_n=0.$$
By the Alternating Series Test $$\sum_{n=1}^{\infty}(-1)^{n+1} a_n $$ converges and we can show that, its sum lies between $a_1$ and $a_1-a_2.$  
For:
Let $S_n=a_1-a_2+a_3-a_4+⋯+(-1)^{n+1} a_n.$
Then
$\forall 2m-1\lt 2n$ we have $$S_{2n}=S_{2m-1}-(a_{2m}-a_{2m+1})-\cdots-(a_{2n-2}-a_{2n-1})-a_{2n}\le S_{2m-1}$$
Also clearly $\forall 2n\lt2m-1$ we have
$$S_{2m-1}=S_{2n}+(a_{2n+1}-a_{2n+2})+\cdots+(a_{2m-3}-a_{2m-2})+a_{2m-1}\ge S_{2n}.$$ This gives us that any odd partial sum is greater than to any even partial sum.
$$S_{2n+1}=S_{2n-1}+(a_{2n+1}-a_{2n} )\le S_{2n-1}$$
$$S_{2n}=S_{2n-2}+(a_{2n-1}-a_{2n} )\ge S_{2n-2}.$$
Therefore
$$S_2\le S_4\le S_6\le\cdots\le S_5 \le S_3 \le S_1$$ and by Monotone Convergent Theorem both limits $$\lim_{n\to\infty}S_{2n}\,\,\,\,\,\text{and}\,\,\,\,\,\,\lim_{n\to\infty}S_{2n+1}$$ exist as real numbers. Also $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}(S_n-S_{n-1})=0.$$ Therefore $$\lim_{n\to\infty}S_{2n}=\lim_{n\to\infty}S_{2n+1}=\lim_{n\to\infty}S_n.$$ Hence $$\sum_{n=1}^{\infty}(-1)^{n+1} a_n $$ converges and $$a_1-a_2\le S_n\le a_1\,\,\,\,\,\forall n\in\mathbb{N}.$$

Hence for any $p\ge 1,$ $$\dfrac{1}{2}\lt\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n^p}\lt1.$$

