Proving the inequality $\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$ I am struggling to prove
$$\prod_{n=1}^k \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$
For all $k \geq 1$. Clearly, this is true for $k=1$, so it suffices to show

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

The obvious way to show this would be to evaluate the product directly, though I don't think that is feasible. I am not sure how else to approach this product. This would be a (much) tighter upper bound than 
$$\exp\left ({\sum_{n=1}^k} \frac1{n^2+\ln n} \right )$$
Which comes from the monotone convergence theorem.
 A: 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.
A: 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.
A: 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 $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$.
