Proving the inequality $\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$

I have not full solution to this problem.

I want to prove by induction that:

$$ \prod_{n=1}^{k} \left(1 + \frac{1}{n^2 + \ln{n}}\right) < \frac{7}{2} \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)} < \frac{7}{2}$$

It's obvious for $k=1$.

Suppose we proved for $k$:

$$ \prod_{n=1}^{k + 1} \left(1 + \frac{1}{n^2 + \ln{n}}\right) < \frac{7}{2} \left(1 + \frac{1}{(k+1)^2 + \ln{(k+1)}}\right) \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)}$$

So we need to prove that:

$$ \left(1 + \frac{1}{(k+1)^2 + \ln{(k+1)}}\right) \frac{(k+1)^2 + \ln(k+1)}{(k+2)^2 + \ln(k+2)} < \frac{(k+2)^2 + \ln(k+2)}{(k+3)^2 + \ln(k+3)} \\ \frac{(k+1)^2 + \ln(k+1) + 1}{(k+2)^2 + \ln(k+2)} < \frac{(k+2)^2 + \ln(k+2)}{(k+3)^2 + \ln(k+3)} $$

But I don't know how to prove the last inequality, which according to Wolfram is true for $k > -1$.

Probably I will improve my answer after a while.

For all $k\in\mathbb{N}$, we have that

$$\prod_{n=1}^k\left(1+\frac{1}{n^2+\log(n)}\right)=\exp\left(\sum_{n=1}^k\log\left(1+\frac{1}{n^2+\log(n)}\right)\right)$$

$$\leq \exp\left(\sum_{n=1}^\infty\log\left(1+\frac{1}{n^2+\log(n)}\right)\right)$$

Let us analyse this infinite sum by using the Taylor series expansion for $\log(1+x)$. We have

$$\log\left(1+\frac{1}{n^2+\log(n)}\right)=\left(\frac{1}{n^2+\log(n)}\right)-\frac{1}{2}\left(\frac{1}{n^2+\log(n)}\right)^2+\cdots $$

$$=-\sum_{m=1}^\infty \frac{(-1)^m}{m}\left(\frac{1}{n^2+\log(n)}\right)^m$$

Then

$$\sum_{n=1}^\infty\log\left(1+\frac{1}{n^2+\log(n)}\right)=-\sum_{n=1}^\infty\sum_{m=1}^\infty \frac{(-1)^m}{m}\left(\frac{1}{n^2+\log(n)}\right)^m$$

$$=\log(2)-\sum_{n=2}^\infty\sum_{m=1}^\infty \frac{(-1)^m}{m}\left(\frac{1}{n^2+\log(n)}\right)^m$$

Since

$$\left|\frac{1}{n^2+\log(n)}\right|<1\text{ for }n>1$$

the sums converge absolutely and we are free to switch the order of summation. Then

$$=\log(2)-\sum_{m=1}^\infty\sum_{n=2}^\infty \frac{(-1)^m}{m}\left(\frac{1}{n^2+\log(n)}\right)^m=\log(2)-\sum_{m=1}^\infty\frac{(-1)^m}{m}\sum_{n=2}^\infty \left(\frac{1}{n^2+\log(n)}\right)^m$$

Note that for all $N\geq 2$, we have that

$$\sum_{n=2}^\infty \left(\frac{1}{n^2+\log(n)}\right)^m\leq \sum_{n=2}^N\left(\frac{1}{n^2+\log(n)}\right)^m+\sum_{n=N+1}^\infty \left(\frac{1}{n^2}\right)^m$$

$$=\sum_{n=2}^N\left(\frac{1}{n^2+\log(n)}\right)^m+\sum_{n=1}^\infty \frac{1}{n^{2m}}-1-\sum_{n=2}^N \frac{1}{n^{2m}}$$

$$=\zeta(2m)-1+\sum_{n=2}^N\left(\left(\frac{1}{n^2+\log(n)}\right)^m-\frac{1}{n^{2m}}\right) $$

We can define

$$f(m,N)=\sum_{n=2}^N\left(\left(\frac{1}{n^2+\log(n)}\right)^m-\frac{1}{n^{2m}}\right)$$

and note that $f(m,N)$ can be explicitly calculated for all $m$ and $N\geq 2$ as it is a finite sum. Then the total sum is

$$\log(2)-\sum_{m=1}^\infty\frac{(-1)^m}{m}\sum_{n=2}^\infty \left(\frac{1}{n^2+\log(n)}\right)^m\leq \log(2)-\sum_{m=1}^\infty\frac{(-1)^m}{m}\left(\zeta(2m)-1+f(m,N)\right)$$

$$=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{m}\left(\zeta(2m)+f(m,N)\right)$$

From our definition of $f(m,N)$, we know that $0<\zeta(2m)+f(m,N)$. Since this is an alternating series with a positive first term, for all $M\in\mathbb{N}$ we have

$$\sum_{m=1}^\infty\frac{(-1)^{m+1}}{m}\left(\zeta(2m)+f(m,N)\right)$$

$$\leq \left|\sum_{m=1}^{M}\frac{(-1)^{m+1}}{m}\left(\zeta(2m)+f(m,N)\right)\right|+\frac{\zeta(2M+2)+f(M+1,N)}{m+1}$$

This can be explicitly calculated as we have removed all infinite series from the equation as the zeta function of even integers is

$$\zeta(2m)=(-1)^{m+1}\frac{B_{2m}(2\pi)^{2m}}{2(2m)!}$$

Here, $B_m$ is the $m$th Bernoulli number and is well known past $m=402$. For $N=M=200$ this comes out to

$$\left|\sum_{m=1}^{200}\frac{(-1)^{m+1}}{m}\left(\zeta(2m)+f(m,200)\right)\right|+\frac{\zeta(402)+f(201,200)}{201}=1.25254$$

Then

$$\exp\left(\sum_{n=1}^\infty\log\left(1+\frac{1}{n^2+\log(n)}\right)\right)\leq e^{1.25254}=3.49923<\frac{7}{2}$$

and the conjecture is proved.

Proof: We will use the following bound: $$\frac{1}{x^2 + \ln x} < \frac{1}{x^2} - \frac{3}{2x^4}, \quad \forall x\ge 5.$$ proof: It suffices to prove that $f(x) = x^2 + \ln x - \frac{2x^4}{2x^2-3} > 0$ for $x\ge 5$. We have $f'(x) = \frac{4x^4+6x^2+9}{x(2x^2-3)^2} > 0$ for $x\ge 5$. Note also that $f(5) > 0$. The desired result follows.

With the bound above, noting also that $\ln (1+x) \le x$ for $x\ge 0$, we have \begin{align} &\prod_{n=1}^\infty \Big(1 + \frac{1}{n^2 + \ln n}\Big)\\ =\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big) \cdot \mathrm{exp}\left(\sum_{n=5}^\infty \ln\Big(1 + \frac{1}{n^2 + \ln n}\Big) \right)\\ \le\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big) \cdot \mathrm{exp}\left(\sum_{n=5}^\infty \Big(\frac{1}{n^2} - \frac{3}{2n^4}\Big) \right)\\ =\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big) \cdot \mathrm{exp}\left(\frac{2689}{13824}+\frac{\pi^2}{6}-\frac{\pi^4}{60} \right)\\ < \frac{7}{2} \end{align} where we have used $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ and $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$.