Prime Number Theorem

Proof. Noted that only prime or prime power will yield non-zero value at von Mangoldt function, which means we must only iterate prime power factors of the input integer, and we have:

In this way, we've shown the connection between von Mangoldt function and natural logarithmic function.

Proof. Again, it's time to utilize the property of to yield non-zero values only at prime or prime power , thus we have:

We would like to show equals asymptotically.

Noted that for every prime prime power , we have , so we can expand the summation ranged from to into to . Similarily, for every prime , there must , and we don't have to constrain to . Thus we have:

While one may suspect we've amplified too much that it must be rather than , but noted that is also a series, and by the rationale above it must also converge to a constant, or . In this way, we can have faith in 's being asymptotically equal to .

Finally we have:

And we are done with the proof.

Proof. Again, due to 's yielding non-zero values at prime or prime power , it's very easy to get to a point that:

The point differs from conclusion we would like to prove only at the upper bound of summation, which is versus . It won't be hard to find when is sufficient large, there will be no summand in . And when there's no summand, must be smaller than the smallest prime , so we have .

Conversely, if we would like to correspond to with non-empty summand, there must be:

And since must be integers, it must be capped at , now we are done.

Proof. First, it can be shown that is bounded above:

By L'Hospital's rules, we can find:

And we know there's trivially , by the squeeze theorem, since is squeezed between and , which tends to as , will also tend to as .

Proof. One can easily do the transformation that:

And it suffices to show:

Proof. First, we need to show #1, let . Noted that , while in the middle of summation, we have:

Where and , so the sequence is non-negative, and thus:

Where the in the summands annihilate to since , which is again due to the hyperbola property of .

Now clearly there's . and we can let , now we have . Consider the following telescopes:

Which can be summed into:

In this way, we've proved #1.

The proof of #2 is just as a simple transformation as:

Where #1 is required to replace into .

Finally, in order to prove #3, we need to interpret the conclusion of #2 as:

Just like what we have done in proving #1, let then we have:

Now we can consider the following inequality:

Note that , which means there definitely exists a chose of such that or , in which we have .

On the other hand, we have:

Which means . So we can let and then complete the proof of #3.

Proof. First, by the Shapiro's tauberian theorem we know , and therefore and exists, we just need to determine their range.

Let , which means , or . Noted that increase monotonically, when we have:

Which means or divide on both sides and we have . The magnitude of can be arbitrarily small by requiring to be sufficiently large so that we have or . And since such must hold for arbitrarily small , it can only be .

Similarily, let , which means , or . And when we have:

And then we have: . The magnitude of can be arbitrarily small by requiring to be sufficiently large so that we have or . And since such must hold for arbitrarily small , it must be .

Combine all of them together and we are done with the proof.

Proof. First, let and compute :

This will lead us to the following formula with multipart summations:

Let's begin with the summation part . First, we need to show there's by L'Hospital's rule:

Which will lead us to:

Here's the craziest part. Noted that each summation parts remained are in the form of generalized convolution, and is the Dirichlet inverse of (by its being completely multiplicative). Let , what we are going to do is looking for ways to represent in the form of . Once it were possible, since they are in the form of , then we will soon have . Of course, error terms might be inevitable and must be handled properly.

By the asymptotic expansion , which can be fixed into , it turns out we can let and give it a shot:

The replacement for is relatively hard to observe (although can be guessed from what we are to prove), instead we will just give out the result for verification:

Thus we can let and give another a shot:

Finally, put all of them back and we will get:

Which is exactly what we are to prove.

Proof. The second part can be obtained directly by expanding , while the first part is due to the Abel's identity:

Where and are used.

Proof. We deliberatedly construct the following expressions:

The error terms can be estimated by:

Thus we have:

Noted that all terms of are positive, by triangle inequality we have:

Which is exactly what we are to prove.

Proof. In the proof of the Abel's identity, noted that the first transformation is completely formal, we just need to fix the parameter of there, which will bring us to:

Which is exactly what we are to prove.

Proof. We would like to let and , then apply the formula above, but let's do some analysis first.

Then, for the difference of , we have:

So by putting them together we have:

We need to handle the stair cases of in and the trailing as the following:

Where the amplification of into is due to , and the substitution of into is due to L'Hospital's rule on .

Thus finally we have:

Which is exactly what we are to prove.

Proof. This is done by a change-of-variable. If we let , which indicates and , then we have:

Proof. Actually one should find no difficulty in establishing and thus it's bounded by constant limit superior , or explicitly .

The term can be controlled to be arbitrarily small (otherwise it's a contradiction to ) by sufficiently large , which is asymptotically equivalent to , and thus can also be interpreted as .

This means at any value is smaller than , the term might be large at small values, but must be controllable by , otherwise it conflicts with the definition of .

Proof. The following property is obvious:

And recall that at every values we have the following inequalities:

So now they can be used to furtherly show the relationship between and that:

Please notice that the properties of and are utilized to bail them out of their own abosolute value brackets, which is due to the amplification of triangle inequalities.

Proof. First, by Shapiro's Tauberian theorem with , we have . And if we take it to the Abel's identity, we have:

Which can be fixed into . Next, subtract from both sides we have:

Finally, let's do a change-of-variable. By letting , which indicates , we have:

Therefore, we have , which indicates , and now we are done.

Proof. By the Selberg's asymptotic formula, we have:

Where the last minification is due to the fact that .

Noted that , which means there's also , and now lies somewhere in between the interval . Consider their distances to midpoint of the interval, we have:

Which can be furtherly fixed into:

Finally, substitute in with , which indicates , noted that , and then we have:

Divides on both sides and then we have:

Where has been used to eliminate .

Proof. Noted that , let , we can see and thus , . While , So we can amplify that:

Now we are done with the proof.

Proof. Recall the form of , where will only change its value at each integer points, and may not change if the current integer point is not prime power, recall that , where is non-zero only if is prime power. Meantime, minimizes the magnitude of as increases, and may create zeroes as drops below . So intuitively, the graph of is in sawtooth shape, with each teeth in the shape of for some .

So the behaviour of are only decreasing continously or increasing by hopping at discontinuities. Since hopping will never decrease its values, to change its sign, it must be negative first, which is due to its being negative first, or has already descended below zero. However, the latter case contradicts with the assumption that there's no zero in the interval. So it must only be the former case, in which either does not change its sign at all or change its sign for at most once.

Proof. The proof is done by assuming and inspecting arbitrary interval with deliberatedly constructed constant stride , defined as below:

Where is the fixed upper bound for we have introduced above.

Consider an arbitrary interval , and in the subinterval , there're three exclusive cases:

1. has at least one zero in , and we let the first zero be .
2. is always positive or always negative.
3. changes its sign for once at .

For the case #1, we have:

Where we let , which fulfils .

For the case #2, we have:

For the case #3, we have:

Both case #2 and #3 can be generalized as the following:

So under any circumstance, there's always:

Finally, for arbitrary , let , and then we have:

Divide on both sides we get:

Take this to the limit superior and then we have:

And thus by assuming we have , which completes the proof.

Proof. Since yields for every , and otherwise. Thus summing them from to yields .

On the other hand, since , by prime number theorem we know , and thus .