I think that the following "derivation" of the Prime Number Theorem from the well-known identity

$\sum_{d|n}\Lambda(d) = \log n$

is a particularly prominent example of what you are asking. Indeed, it follows from the said identity that

$\sum_{n \leq x} \sum_{d|n} \Lambda(n) = \sum_{d\leq x} \Lambda(d)\sum_{n \leq x, d|n} 1 = \sum_{d\leq x}\Lambda(d)\lfloor \frac{x}{d}\rfloor$

and whence,

$\sum_{n \leq x}\Lambda(n)\lfloor \frac{x}{n} \rfloor = \sum_{n \leq x} \log n \sim x \log x$.

Now if we replaced the $\lfloor x/n \rfloor$ in the previous line by $x/n$, we would get

$\sum_{n\leq x}\frac{\Lambda(n)}{n} \sim \log x \sim \sum_{n \leq x}\frac{1}{n}$.

This might lead us to ascertain that the function $\Lambda$ of von Mangoldt behaves in the average like the arithmetical function that is identically equal to $1$, thus

$\psi(x):=\sum_{n\leq x} \Lambda(n)\sim x.$ (The Prime Number TheoremVoilà!)

I think that the following "derivation" of the Prime Number Theorem from the well-known identity

$\sum_{d|n}\Lambda(d) = \log n$

is a particularly prominent example of what you are asking. Indeed, it follows from the said identity that

$\sum_{n \leq x} \sum_{d|n} \Lambda(n) = \sum_{d\leq x} \Lambda(d)\sum_{n \leq x, d|n} 1 = \sum_{d\leq x}\Lambda(d)\lfloor \frac{x}{d}\rfloor$

and whence,

$\sum_{n \leq x}\Lambda(n)\lfloor \frac{x}{n} \rfloor = \sum_{n \leq x} \log n \sim x \log x$.

Now if we replaced the $\lfloor x/n \rfloor$ in the previous line by $x/n$, we would get

$\sum_{n\leq x}\frac{\Lambda(n)}{n} \sim \log x \sim \sum_{n \leq x}\frac{1}{n}$.

This might lead us to ascertain that the function $\Lambda$ of von Mangoldt behaves in the average like the arithmetical function that is identically equal to $1$, thus

$\psi(x):=\sum_{n\leq x} \Lambda(n)\sim x.$ (The Prime Number Theorem!)

As to the rigorous formal version of the preceding argument you may want to take a look at sections 9.9 through 9.12 of [2]. You are to find there a proof of the Prime Number Theorem (presumably due to Ingham) based on the estimate

$\sum_{n \leq x} \psi(\frac{x}{n}) = x\log x - x+ O(\log x), \quad x \geq 1.$

According to Prof. Balanzario (see [1, page 59]): "This demonstration ... is the correct version of our heuristic reasoning [given above]."

References

[1] E. P. Balanzario. Breviario de Teoría Analítica de los Números. SMM, México, 2003.

[2] W. Rudin. Functional Analysis. Tata McGraw Hill Publishing Company Ltd., 1974.

I think that the following "derivation" of the Prime Number Theorem from the well-known identity

$\sum_{d|n}\Lambda(d) = \log n$

is a particularly prominent example of what you are asking. Indeed, it follows from the said identity that

$\sum_{n \leq x} \sum_{d|n} \Lambda(n) = \sum_{d\leq x} \Lambda(d)\sum_{n \leq x, d|n} 1 = \sum_{d\leq x}\Lambda(d)\lfloor \frac{x}{d}\rfloor$

and whence,

$\sum_{n \leq x}\Lambda(n)\lfloor \frac{x}{n} \rfloor = \sum_{n \leq x} \log n \sim x \log x$.

Now if we replaced the $\lfloor x/n \rfloor$ in the previous line by $x/n$, we would get

$\sum_{n\leq x}\frac{\Lambda(n)}{n} \sim \log x \sim \sum_{n \leq x}\frac{1}{n}$.

This might lead us to ascertain that the function $\Lambda$ of von Mangoldt behaves in the average like the arithmetical function that is identically equal to $1$, thus

$\psi(x):=\sum_{n\leq x} \Lambda(n)\sim x.$ (The Prime Number Theorem!)

As to the rigorous version of the preceding argument you may want to take a look at sections 9.9 through 9.12 of [2]. You are to find there a proof of the Prime Number Theorem (presumably due to Ingham) based on the estimate

$\sum_{n \leq x} \psi(\frac{x}{n}) = x\log x - x+ O(\log x), \quad x \geq 1.$

Let me co

References

[1] E. P. Balanzario. Breviario de Teoría Analítica de los Números. SMM, México, 2003.

[2] W. Rudin. Functional Analysis. Tata McGraw Hill Publishing Company Ltd., 1974.

I think that the following "derivation" of the Prime Number Theorem from the well-known identity

$\sum_{d|n}\Lambda(d) = \log n$

is a particularly prominent example of what you are asking. Indeed, it follows from the said identity that

$\sum_{n \leq x} \sum_{d|n} \Lambda(n) = \sum_{d\leq x} \Lambda(d)\sum_{n \leq x, d|n} 1 = \sum_{d\leq x}\Lambda(d)\lfloor \frac{x}{d}\rfloor$

and whence,

$\sum_{n \leq x}\Lambda(n)\lfloor \frac{x}{n} \rfloor = \sum_{n \leq x} \log n \sim x \log x$.

Now if we replaced the $\lfloor x/n \rfloor$ in the previous line by $x/n$, we would get

$\sum_{n\leq x}\frac{\Lambda(n)}{n} \sim \log x \sim \sum_{n \leq x}\frac{1}{n}$.

This might lead us to ascertain that the function $\Lambda$ of von Mangoldt behaves in the average like the arithmetical function that is identically equal to $1$, thus

$\psi(x):=\sum_{n\leq x} \Lambda(n)\sim x.$

As to the rigorous version of the preceding argument you may want to take a look at sections 9.9 through 9.12 of [2]. You are to find there a proof of the Prime Number Theorem (presumably due to Ingham) based on the estimate

$\sum_{n \leq x} \psi(\frac{x}{n}) = x\log x - x+ O(\log x), \quad x \geq 1.$

Let me co

References

[1] E. P. Balanzario. Breviario de Teoría Analítica de los Números. SMM, México, 2003.

[2] W. Rudin. Functional Analysis. Tata McGraw Hill Publishing Company Ltd., 1974.

I think that the following "derivation" of the Prime Number Theorem from the well-known identity

$\sum_{d|n}\Lambda(d) = \log n$

is a prominent example. Indeed, it follows from the said identity that

$\sum_{n \leq x} \sum_{d|n} \Lambda(n) = \sum_{d\leq x} \Lambda(d)\sum_{n \leq x, d|n} 1 = \sum_{d\leq x}\Lambda(d)\lfloor \frac{x}{d}\rfloor$

and whence,

$\sum_{n \leq x}\Lambda(n)\lfloor \frac{x}{n} \rfloor = \sum_{n \leq x} \log n \sim x \log x$.

Now if we replaced the $\lfloor x/n \rfloor$ in the previous line by $x/n$, we would get

$\sum_{n\leq x}\frac{\Lambda(n)}{n} \sim \log x \sim \sum_{n \leq x}\frac{1}{n}$.

This might lead us to ascertain that the function $\Lambda$ of von Mangoldt behaves in the average like the arithmetical function that is identically equal to $1$, that's to saythus

$\psi(x):=\sum_{n\leq x} \Lambda(n)\sim x.$

As to the rigorous version of the preceding argument you may want to take a look at sections 9.9 through 9.12 of [2]. 2]. You are to find there a proof of the Prime Number Theorem (presumably due to Ingham) based on the estimate

$\sum_{n \leq x} \psi(\frac{x}{n}) = x\log x - x+ O(\log x), \quad x \geq 1.$

References

[1] 1] E. P. Balanzario. Breviario de Teoría Analítica de los Números. SMM, México, 2003.

[2] 2] W. Rudin. Functional Analysis. Tata McGraw Hill Publishing Company Ltd., 1974.

I think that the following "derivation" of the Prime Number Theorem from the well-known identity

$\sum_{d|n}\Lambda(d) = \log n$

is a prominent example. Indeed, it follows from the said identity that

$\sum_{n \leq x} \sum_{d|n} \Lambda(n) = \sum_{d\leq x} \Lambda(d)\sum_{n \leq x, d|n} 1 = \sum_{d\leq x}\Lambda(d)\lfloor \frac{x}{d}\rfloor$

and whence,

$\sum_{n \leq x}\Lambda(n)\lfloor \frac{x}{n} \rfloor = \sum_{n \leq x} \log n \sim x \log x$.

Now if we replaced the $\lfloor x/n \rfloor$ in the previous line by $x/n$, we would get

$\sum_{n\leq x}\frac{\Lambda(n)}{n} \sim \log x \sim \sum_{n \leq x}\frac{1}{n}$.

This might lead us to ascertain that the function $\Lambda$ of von Mangoldt behaves in the average like the arithmetical function that is identically equal to $1$, that's to say

$\psi(x):=\sum_{n\leq x} \Lambda(n)\sim x.$

As to the rigorous version of the preceding argument you may want to take a look at sections 9.9 through 9.12 of [2]. You are to find there a proof of the Prime Number Theorem based on the estimate

$\sum_{n \leq x} \psi(\frac{x}{n}) = x\log x - x+ O(\log x), \quad x \geq 1.$

References

[1] E. P. Balanzario. Breviario de Teoría Analítica de los Números. SMM, México, 2003.

[2] W. Rudin. Functional Analysis. Tata McGraw Hill Publishing Company Ltd., 1974.