Question

Solve for all N > 2.

$\sqrt{2{\sqrt{3{\sqrt{4{\sqrt{...{\sqrt{N-1}}}}}} < 3$ ​

Answer 1

Step-by-step explanation:

This is a decent (and probably long) exercise.

For some n∈N , we want to show that

Fn=234⋯(n−1)n−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−⎷<3

(Denote by Fn the value of this expression)

We begin by observing that this can be rewritten as the finite product

Fn=1⋅212⋅3122⋅4123⋯n12n−1

=∏k=0n−1(k+1)12k

Here, an important realisation is that (k+1)12k>1 for all k≥1 , then it might help if we consider the infinite product of Fn since

∏k=0n−1(k+1)12k<∏k=0∞(k+1)12k

Now, we will show that F∞<3 . This in turn, would imply Fn<3 for all n∈N . Though before this, we simplify the product F∞ by writing it as an infinite series via taking natural logarithm both sides. This yields

lnF∞=∑k=0∞ln(1+k)2k =∑k=1∞ln(1+k)2k

Here, the main idea is to find a strong upper bound for this tedious sum. To do this, we consider another inequality (which will immediately become relevant to our case). For all n∈N , we have

n3≤3n

(with the equality holding only if n=3 )

This can be easily shown by induction (by noting that

Answer 2

Answer:

The main changes were:

This is a decent (and probably long) exercise.

For some n∈Nn∈N , we want to show that

Fn=234⋯(n−1)n−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−⎷<3Fn=234⋯(n−1)n<3

(Denote by FnFn the value of this expression)

We begin by observing that this can be rewritten as the finite product

Fn=1⋅212⋅3122⋅4123⋯n12n−1Fn=1⋅212⋅3122⋅4123⋯n12n−1

=∏k=0n−1(k+1)12k =∏k=0n−1(k+1)12k

Here, an important realisation is that (k+1)12k>1(k+1)12k>1 for all k≥1k≥1 , then it might help if we consider the infinite product of FnFn since

∏k=0n−1(k+1)12k<∏k=0∞(k+1)12k∏k=0n−1(k+1)12k<∏k=0∞(k+1)12k

Now, we will show that F∞<3F∞<3 . This in turn, would imply Fn<3Fn<3 for all n∈Nn∈N . Though before this, we simplify the product F∞F∞ by writing it as an infinite series via taking natural logarithm both sides. This yields

lnF∞=∑k=0∞ln(1+k)2k =∑k=1∞ln(1+k)2kln⁡F∞=∑k=0∞ln⁡(1+k)2k =∑k=1∞ln⁡(1+k)2k

Here, the main idea is to find a strong upper bound for this tedious sum. To do this, we consider another inequality (which will immediately become relevant to our case). For all n∈Nn∈N , we have

n3≤3nn3≤3n

(with the equality holding only if n=3n=3 )

This can be easily shown by induction (by noting that

(n+1)3=n3(1+1n)3<3n3<3⋅3n=3n+1(n+1)3=n3(1+1n)3<3n3<3⋅3n=3n+1

for all n≥4n≥4 since (1+1n)3(1+1n)3 is decreasing on NN and this holds for the base case n=4n=4 ).

We manipulate this to use it in this case.

n≤(313)nn≤(313)n

Further, by taking natural logarithm both sides, we obtain

lnn≤ln33⋅nln⁡n≤ln⁡33⋅n

for all n∈Nn∈N . Using this inequality, we obtain the upper bound for our sum. This is given by

∑k=1∞ln(1+k)2k<ln33∑k=1∞1+k2k∑k=1∞ln⁡(1+k)2k<ln⁡33∑k=1∞1+k2k

(we have a strict inequality since the equality of the previous result holds only if n=3n=3 )

From here, computing the exact value of the upper bound of our sum is straightforward. The first sum on the right side is an infinite geometric series.

∑k=1∞12k=121−12=1∑k=1∞12k=121−12=1

For the second sum, we consider the infinite geometric series:

∑k=0∞xk=11−x∑k=0∞xk=11−x

(which is convergent for |x|<1|x|<1 )

Here, we differentiate both sides with respect to xx and then multiply by xx on both sides. This yields

∑k=1∞k⋅xk=x(1−x)2∑k=1∞k⋅xk=x(1−x)2

With x=12x=12 , the second sum comes out to be

∑k=1∞k2k=12(1−12)2=2∑k=1∞k2k=12(1−12)2=2

Putting it all together, we have shown that

∑k=1∞ln(1+k)2k<ln33(1+2)=ln3∑k=1∞ln⁡(1+k)2k<ln⁡33(1+2)=ln⁡3

Since the left side of this inequality is lnF∞ln⁡F∞ , we take exponents both sides to conclude that

F∞<3F∞<3

which combines with the earlier statement Fn<F∞Fn<F∞ to provide the desired inequality.

Fn=234⋯(n−1)n−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−