Establish a lower bound for the generalized alternating harmonic series

Question

I'm given the following series:$$\sum_{n=1}^{\infty}{(-1)^{n-1}\over n^p}$$ I need to show that the sum is greater than $1/2$ for every $p > 0$. For $p \ge1$ this is obvious, as it follows by grouping terms $2$ by $2$, and the first sum is greater than $1/2$ while the other terms are all positive. I'm stuck with $0<p<1$.

Answer 1

let $S_p(N)=\sum_{k=1}^{N}{k^{-p}}$; for notational simplicity we fix for now $0<p<1$ and let $S_p(N)=S(N)$.

We note that $\eta(p)=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n^p} > S(2N)-2^{1-p}S(N)$ for any $N \ge 1$ as the remainder is an alternating sum with decreasing terms that starts at the positive $\frac{1}{(2N+1)^p}$

Using that $f(x)=x^{-p}$ is convex as $f''(x)>0, x>0$, we get $f(k) \le \int_{k-\frac{1}{2}}^{k+\frac{1}{2}}f(x)dx$ since $f(k+\alpha)+f(k-\alpha) \ge 2f(k), k \ge 1, 0 \le \alpha \le \frac{1}{2}$. Hence $S(N) \le \int_{\frac{1}{2}}^{N+\frac{1}{2}}f(x)dx=\frac{(N+\frac{1}{2})^{1-p}-\frac{1}{2}^{1-p}}{1-p}$, or

$-2^{1-p}S(N) \ge -\frac{(2N+1)^{1-p}}{1-p}+\frac{1}{1-p}$

Using the trapezoidal rule for $f$ which is convex, so the error term is negative, we get $\int_2^{2N}f(x)dx \le f(2)+..f(2N)-\frac{1}{2}f(2)+O(N^{-p})$, or

$S(2N) \ge \frac{(2N)^{1-p}}{1-p}+1+\frac{1}{2^{p+1}}-\frac{2^{1-p}}{1-p}+O(N^{-p})$

Hence $\eta(p) > \frac{(2N)^{1-p}}{1-p}+1+\frac{1}{2^{p+1}}-\frac{2^{1-p}}{1-p}+O(N^{-p})-\frac{(2N+1)^{1-p}}{1-p}+\frac{1}{1-p}$. Since the sum of terms in $N$ obviously goes to zero when $0<p<1$ fixed, it is enough to prove that

$1+\frac{1}{2^{p+1}}-\frac{2^{1-p}}{1-p}+\frac{1}{1-p}>\frac{1}{2}$ as then we pick $N$ large enough st the terms in $N$ are less than half $1+\frac{1}{2^{p+1}}-\frac{2^{1-p}}{1-p}+\frac{1}{1-p}-\frac{1}{2}>0$ and we get $\eta(p) > \frac{1}{2}$ as required

By bringing to the same denominator and allowing $0 \le p \le 1$ the required inequality is equivalent to:

edit - as pointed out the inequality is slightly more complicated but still elementary as follows:

$(3-p)2^p \ge 3+p$ with equality only for $p=0,1$

Here we let $q=2^p, p\log 2 =\log q, 1 \le q \le 2$, so we consider $g(q)=(3\log 2-\log q)q-3\log 2-\log q$ and we need to show that $g(q) >0, 1<q<2$ But $g''(q)=\frac{1}{q^2}-\frac{1}{q} \le 0$ as $q \ge 1$ so $g'$ decreasing and since $g'(1)=3\log 2-2>0, g'(2)=2\log 2-1.5<0$, it follows that $g$ strictly increases until some $1<q_0<2$ and then strictly decreases to $g(2)$, while $g(1)=g(2)=0$ ensuring $g(q)>0, 1<q<2$, so done!

Note that if we only require $\eta(p) \ge \frac{1}{2}$ we can directly apply the trapezoidal rule from $1$ to $2N$ rather than $2$ to $2N$ and then we get precisely $S(2N)-2^{1-p}S(N) \ge \frac{1}{2}-c_N$ where $|c_N| \to 0$ when $N \to \infty$, hence the result follows. To get a strict inequality (which is true as we saw) we need the extra work and that way we can get better estimates for $p$ away from $0$ or $1$