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$\sum_{n=1}^{\infty} \frac{\varphi(n)}{n}$ where $\varphi(n)$ is 1 if the variable $\text n$ has the number $\text 7$ in its typical base-$\text10$ representation, and $\text0$ otherwise.

I am supposed to find out if this series converges or diverges. I think it diverges, and here is why.

We can see that there is a series whose partial sums are always below our series, but which diverges. Compare some of the terms of each sequence

$\frac{1}{7} > \frac{1}{8}$
$\frac{1}{70} > \frac{1}{80}$
$\frac{1}{71} > \frac{1}{80}$
$\frac{1}{72} > \frac{1}{80}$
$\text ... $
$\frac{1}{79} > \frac{1}{80}$
$\text ... $
$\frac{1}{700} > \frac{1}{800}$
$\text ... $

And continue in this way.

Obviously some terms are left out of the sequence on the left, which is fine since our sequence of terms on the left is already greater than the right side. Notice the right side can be grouped into

$\frac{1}{8} + \frac{1}{8} + ... $ because we will have $10$ $\frac{1}{80}$s, $100$ $\frac{1}{800}$s, etc etc. Thus we are adding up infinitely many 1/8s. This is similar to the idea of the divergence of the harmonic series. So, my conclusion is that it diverges. A bunch of other students in my real analysis class have come to the conclusion that is, in fact, convergent, and launched into a detailed verbal explanation about comparison with a geometric series that I couldn't follow without seeing their work. Is my reasoning, like they suspect, flawed? I can't see how.

Sorry about the poor format, I'm new to TeX and couldn't figure out how to format a piecewise function (it was telling me a my \left delimiter wasn't recognized).

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The series doesn't make sense. What is $i$? Why would a finite sum diverge? –  Asaf Karagila Oct 2 '10 at 22:24
    
Nothing to worry about, anon: in "\sum_{i=1}^{10} \frac{\varphi(n)}{n}", just change 10 to \infty. –  Vandermonde Oct 2 '10 at 22:26
    
how embarrassing, it is from n = 1 to infinity. I edited it. –  anon Oct 2 '10 at 22:31
    
Testing $ \sum_{i=1}^{\infty} \frac{\varphi(n)}{n} $ instead of $ \sum_{n=1}^{\infty} \frac{\varphi(n)}{n} $ would certainly be easier, though! –  Vandermonde Oct 2 '10 at 22:36
    
I see no problem with your reasoning and agree that the series diverges. –  aschepler Oct 2 '10 at 23:14
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3 Answers

up vote 4 down vote accepted

Yes your argument seems fine.

Another argument is:

Divide the integers into blocks of 10

$$[1 \dots 10] [11 \dots 20] [21 \dots 30] \dots$$

Each block will have an number with digit $7$ in it and so the series has sum at least

$$\frac{1}{10} + \frac{1}{20} + \dots + \frac{1}{10n} + \dots $$

$$ = \frac{1}{10}(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} + \dots)$$

which is divergent.

Yet another argument which uses the following useful result (the main reason for posting my answer):

If $A = \{a_i\}$ is a sequence of natural numbers such that $\sum_{i=1}^{\infty} \frac{1}{a_i}$ converges then the natural density of $A$ is zero. (This is an exercise in Ivan Niven's book on Number theory and for a proof see here: http://math.stackexchange.com/questions/5932/theorem-on-natural-density/5967#5967).

In our case the density of the numbers under consideration is at least $\frac{1}{10}$ and thus the sum cannot be convergent.

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Your argument seems fine to me... so I'll give the argument that popped into my head when I read the question.

Sum phi(n)/n for n congruent to 7 (mod 10) and multiply by 10 (which does not affect divergence/convergence). Note that

10/7  > 1/7 +1/8 +...+1/16
10/17 > 1/17+1/18+...+1/26

and so on. The result is "larger" than the harmonic series minus the first six terms (which diverges).

Although, I actually prefer the OP's proof since it's self-contained (i.e. doesn't require the harmonic series).

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+1: I suppose my first argument is very similar to yours. –  Aryabhata Oct 3 '10 at 2:25
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Here's another proof that your sum diverges.

Consider the sum $\sum_{n-1}^\infty (1-\phi(n))/n$. This is the sum of the reciprocals of integers which don't have a 7 in their decimal expansion.

The number of integers $n$ with $1-\phi(n)=1$ and $1 \le n < 10^k$ is $9^k - 1$. (We can choose each of $k$ decimal digits to be anything but 7, except we don't want to choose all zeroes.) Thus the numbers of integers $n$ with $1-\phi(n) = 1$ and $10^{k-1} \le n < 10^k$ is $(9^k - 1) - (9^{k-1}-1) = 8 \times 9^{k-1}$.

Therefore $$ S_k = \sum_{n=10^{k-1}}^{10^k - 1} {1-\phi(n) \over n} $$ has $8 \times 9^{k-1}$ nonzero terms; they are each at most $10^{k-1}$. So $S_k \le 8 \times (9/10)^{k-1}$.

The infinite sum is then $$ \sum_{k=1}^\infty S_k \le 8 \times \sum_{k=1}^\infty (9/10)^{k-1} = 8 \times 10 = 80 $$ and in particular it's finite.

Now, the harmonic series $\sum_{n=1}^\infty 1/n$ diverges; removing terms whose sum converges (like the sum that I just showed to converge) won't change that. So your series diverges.

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I had a professor give this series as an example in an undergraduate real analysis lecture. Looking at the first several terms gives the impression that you're basically dealing with the harmonic series minus a few terms. But he diffused that idea quickly by beginning his his explanation with "Most integers have more than a billion digits." –  I. J. Kennedy Nov 17 '10 at 23:49
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