Let $f(k)=2^k$ and if the value of $\lim_{n\to \infty}16^{f(n)}\overset{k=n}{\underset{k=1}{\prod}}\frac{1}{(2^{2f(k)}-2^{f(k)}+1)}$ is P then find P/7.
This is how I tried to solve:
$$P=\lim_{n\to \infty}16^{f(n)}\overset{k=n}{\underset{k=1}{\prod}}\frac{1}{(2^{2f(k)}-2^{f(k)}+1)}$$
Now if we somehow convert the product into summation then we can apply limit and convert it into integral form and then it becomes easy to evaluate the integral.
So it took the natural logarithm of P.
$$\ln P=\ln \left(\lim_{n\to \infty}16^{f(n)}\overset{k=n}{\underset{k=1}{\prod}}\frac{1}{(2^{2f(k)}-2^{f(k)}+1)}\right)$$
$$\implies \ln P=\lim_{n\to \infty}\ln \left(16^{f(n)}\overset{k=n}{\underset{k=1}{\prod}}\frac{1}{(2^{2f(k)}-2^{f(k)}+1)}\right)$$
As we know that $\ln \left(\overset{n}{\underset{k=1}{\Pi}}k\right)=\ln \left(1.2.3.....n\right)=\ln 1 +\ln 2+ \ln 3 +...\ln n $
$\implies \ln \left(\overset{n}{\underset{k=1}{\Pi}}k\right)=\sum_{k=1}^n\ln k$
So we can write P as follow:
$$\ln P=\lim_{n\to \infty}\sum_{k=1}^{n}\ln \left(\frac{16^{f(n)}}{(2^{2f(k)}-2^{f(k)}+1) }\right)$$
But here I am stuck as I am unable to manipulate the term $\frac{16^{f(n)}}{(2^{2f(k)}-2^{f(k)}+1) }$ so that the above limit can be converted into integral form. I tried a lot but no ideas.If the above method does not work then what are the other ways for evaluating the limit. Any help will be appreciated.
The value of P/7 is given as 3.