1. Let $P_n=\prod_{k=1}^{n}\left(2^{1/2}-2^{\frac{1}{2k+1}}\right).$ We need to find $\lim_{n\to\infty} P_n.$

2. Observe the behavior of each factor. For every $k\ge 1$, $\frac{1}{2k+1}<\frac12.$ Since the function $2^x$ is increasing, $2^{\frac{1}{2k+1}}<2^{1/2}=\sqrt{2}.$ Hence each factor is positive: $0<\sqrt{2}-2^{\frac{1}{2k+1}}.$

Also, $\sqrt{2}-2^{\frac{1}{2k+1}}<\sqrt{2}-1<1,$ because $2^{\frac{1}{2k+1}}>1$. So every factor lies in $(0,1)$.

3. In particular, the very first factor is $\sqrt{2}-2^{1/3}.$ Numerically, $\sqrt{2}\approx 1.414,\qquad 2^{1/3}\approx 1.260,$ so $0<\sqrt{2}-2^{1/3}\approx 0.154<1.$

Thus, $0<P_n\le \left(\sqrt{2}-2^{1/3}\right)^n?$ This inequality is not directly valid because factors are not all equal, so instead we use a simpler argument: Since all factors are positive and each is strictly less than $1$, the sequence $P_n$ is decreasing and bounded below by $0$.

4. Now examine the factors for large $k$. As $k\to\infty$, $2^{\frac{1}{2k+1}}\to 2^0=1.$ Therefore, $\sqrt{2}-2^{\frac{1}{2k+1}}\to \sqrt{2}-1.$ And $0<\sqrt{2}-1<1.$

So for sufficiently large $k$, each factor is less than some constant $r$ with $\sqrt{2}-1<r<1.$ For example, choose any such $r$; then for all large $k$, $\sqrt{2}-2^{\frac{1}{2k+1}}<r.$ Hence the tail of the product is bounded by a geometric-type product: $P_n\le C\, r^{\,n-N}$ for some constant $C>0$ and some fixed $N$. Since $0<r<1$, $r^{n-N}\to 0.$ Therefore, $P_n\to 0.$

5. Hence, $\boxed{\lim_{n\to\infty}\prod_{k=1}^{n}\left(2^{1/2}-2^{\frac{1}{2k+1}}\right)=0.}$

6. Checking options:

• A: $\sqrt2$ — incorrect
• B: $1$ — incorrect
• C: $\frac1{\sqrt2}$ — incorrect
• D: $0$ — correct