I am trying to prove that the following inequality holds $\forall k \in \mathbb{N} \geq 8$:
$$\sum_{n=1}^{\infty} \frac{1}{n} \left( k-2n+\frac{(2n)^{k+1}}{(2n+1)^{k}} \right) > k \left(\ln (\pi)+ \gamma\right)$$
where $\gamma$ denotes the Euler-Mascheroni constant.
I have tried to use induction, but have gotten stuck with the sum, as I'm not sure what to do with the $\frac{(2n) (2n)^{k+1}}{(2n+1)(2n+1)^k}$ section. I have also tried to use an integral test for convergence on the sum in order to bound it, specifically:
$$\int_{1}^{\infty} f(x, k) \, dx\leq \sum_{n=1}^{\infty} f(n, k) \leq f(1, k) + \int_{1}^{\infty} f(x,k) \, dx$$
Whilst this attempt held promise, Mathematica gave the lower bound integral to be:
$$\int_{1}^{\infty} f(x,k) \, dx= k H_k-k (1+\ln (2))+(-1)^{-k} B_{-2}(k+1,1-k)+2$$
where $H_k$ denotes the $k$-th harmonic number, and $B_z$ denotes the incomplete beta function. It is not clear to me if this integral is always greater than $k \left( \ln (\pi) + \gamma\right)$.
I have also attempted to bound the sum with some other sum that I know is bigger than $k \left( \ln (\pi) + \gamma\right)$, namely:
$$k \left(\ln (\pi)+ \gamma\right)<\sum_{n=1}^{\infty} \frac{(k+1)}{n!} < \sum_{n=1}^{\infty} \frac{1}{n} \left( k-2n+\frac{(2n)^{k+1}}{(2n+1)^{k}} \right)$$
As $\sum_{n=1}^{\infty} \frac{(k+1)}{n!} = (k+1)(e-1)$, this implies that $1+\frac{1}{k} > \frac{\ln(\pi) + \gamma}{e-1} \approx 0.76437$, which is definitely true for any positive integer $k$. However, this approach only works for $k \geq 10$, and I'm not sure how to show that it always less than the sum I'm actually interested in- I have tried asymptotics, but haven't gotten anywhere far.
I am wondering if there is any other elegant approach that I am not currently thinking of to prove the inequality at the beginning.
Best Answer
Since $(-1)^kB_{-2}(k+1,1-k)=-\int_0^2u^k(1+u)^{-k}\,du>-2$ we have $\int_1^\infty f(x,k)\,dx=k(H_k-1-\log2)>k(\log\pi+\gamma)$ for all $k>16$.
The range $8\le k\le15$ can be checked computationally.