**1 History**

In 1874, Mertens proved that

where is Euler’s totient function. A natural question we can ask is whether or not this error term can be improved, and what we should expect it to look like. Throughout we use the notation for the totient summatory function, and for the error function. Our goal is to explore the known results for this error term, and in particular the proof of the best known lower bound due to Montgomery. The best unconditional upper bound for was given by Walfisz in 1963 using exponential sums:

Prior to this, in 1930, Chowla and Pillai showed that the error term cannot be improved too much as that is **not** . In particular they looked at the average of and proved that

This tells us that on average, and some more work can be done to gain the factor of . In 1950, Erdos and Shapiro proved that it alternates sign, that there exists such that for infinitely many integers we have

or more concisely, using big- notation,

In 1987 Montgomery improved this to

Moreover, Montgomery conjectured that both , and hold. In what follows, we will examine Montgomery’s proof in more detail.

**2 Preliminaries: Mertens Result and Restating the Question**

We can use the multiplicative properties of the function to obtain the identity where is the mobius function. Using this, and then rearranging the order, we may write

Taking out the factor of and using Gauss’s identity the right hand side becomes

The rightmost sum is exactly since for and Hence

Writing the floor function as where is the fractional part, we see that

Using the fact that and the first and last term may be rewritten as

(Remark: We can actually get an error term of with more work. Since it will not affect anything, we continue with ). Putting this together with the definition of we see that

and the bound implies Merten’s Result

The main goal of this section was to show that we can completely understand by understanding the correlation between the mobius function and the fractional part in the sum Using the prime number theorem, that is the result partial summation tells us that which allows us to rewrite our identity as

where is the sawtooth function. We also define

and will deal with this function to avoid the extraneous factor of

**3 Montgomery’s Lower Bound**

Chowla’s proof that was centered around evaluated the average value of In a similar way, Montgomery’s proof is based on looking at the average of over an arithmetic progression. That is, the goal will be to show that given $latedx x$ , for some suitable we have

and similarly for the negative value. From this it follows that for some will be greater then or less then the average, implying the desired result about

An intuitive reason why we expect to be able to manipulate the average of over an arithmetic progression is the identity for a sum of the sawtooth function over a progression:

If I sum over an arithmetic progression, and then switch the orders of summation, this identity gives us a hope that we will be able to deal with the inner sum comfortably. Exactly this approach is facilitated by the following lemma which allows us to shift and line up the end of the inner sum’s before switching the order, and without creating a large error.

**Lemma** We have

uniformly for and

*Proof Sketch.* We largely skip the analytic details. This is straightforward and follows from using the fact that and then bounding the various parts of the sum.

We will now try switching, and simplifying the inner sum. By the lemma, we see that for

Using the periodicity of the sawtooth function, and the identity we have that

Hence

In what follows the modulus cannot be too large to deal with the error term, so we take and then specify For the sum in the main term, we partition based on the values of Since we have a factor of assume that is squarefree and write with and so that Then

Removing the condition we introduce and error term

where the last follows from the upper bound condition on Thus we have proven

**Theorem**Uniformly for we have

where

Note that this estimate is nontrivial since can grow with as can grow with Also, it is worth mentioning that the error term can be improved from to by modifying various steps throughout, however that will not play a role in the computations which follow.

Making the correct choices for and will lead to a proof of the result. The goal is then to choose and such that is maximized. Montgomery does this by exploiting the primes which are non squares for a particular modulus. If I take an odd number of non-squares, the product is a non square, and if I take an even number, the product is a square. This gives a correlation with the Mobius function since an odd number of factors returns -1 , and an even number of factors returns 1 . Working modulo 4 , let

We choose so that and choose whether or not to remove the prime 3 from the product as to force to be even. Then let so that we are trying to understand the sum

If is odd, then and then Similarly, if is even, Thus we are looking at

The sum can be evaluated to be approximately by using facts about numbers composed only of primes in a particular arithmetic progression. If we instead chose the sum reverses sign. That is From this, we have to conclude that there exists with

Since this factor can be discarded because of the \log\log x term, and we have proven that

which implies the desired result