The limit at infinity of a multiplicative arithmetic function I need to prove this result:
Let $f$ be a multilicative arithmetic function. With p prime, if $ \displaystyle\lim_{p^m \to{+}\infty}{f(p^m)}=0$ then $ \displaystyle\lim_{n \to{+}\infty}{f(n)}=0$
The thing is, I really don't know if this is true. Any help?
 A: The statement is false, even if you restrict yourself to completely multiplicative arithmetic functions. 
Let $f(n): \mathbb{N} \to \mathbb{Q}$ be a function defined by $f(n)=\frac{1}{2^e}$, where 
$$n=p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}$$
is a factorization of $n$ into prime powers (i.e., each $p_i$ is a prime, $p_i\neq p_j$ unless $i=j$, and $e_i\geq 1$), and put $e=\sum_{i=1}^r e_i$. In other words, $f(p^e)=\frac{1}{2^e}$ and we extend this to a multiplicative function on $\mathbb{N}$.
Then 


*

*$f(n)$ is completely multiplicative,

*$f(p^e)=\frac{1}{2^e}\to 0$ as $e\to \infty$, and

*$f(n)=\frac{1}{2}$ infinitely often (for each $n$ prime, and there are infinitely many of those!).
Hence $\lim_{n\to \infty} f(n)\neq 0$. In fact the limit does not exist.
A: The statement is false.  Suppose $f(p^m)=\begin{cases}p & m=1\\ 0 & m>1\end{cases}$.  For any fixed prime the limit is 0 as soon as $m>1$, however as $n\rightarrow \infty$ you keep getting primes and so the limit is not 0.
A: The function 
$$f(n)=\begin{cases}1&\text{if }n\text{ odd}\\0&\text{if }n\text{ even}\end{cases}$$
is multiplicative and we have $\lim_{m\to\infty}f(p^m)=0$ if we let $p=2$. Since $\lim_{n\to\infty}f(n)$ does not exist, I guess that we may assume that for all primes $p$ we have $\lim_{m\to\infty}f(p^m)=0$.
But this is still not good enough:
A multiplicative function is uniqeuly determined by its values at prime powers. Even if we assume that $f$ should be strongly mutiplicative, we can let $f(p^m)=(1-\frac1p)^m$. Then indeed $f(p^m)\to0$ as $m\to\infty$ for any given $p$ (in fact, for any integer $p>1$), but since arbitraryly big primes exist, we have arbitrarily big numbers $n=p$ with $f(n)=(1-\frac1p)>\frac12$.
In summary: I cannot find an interpretation of the problem statement that would make the claim true.
A: What you need to prove is true. The problem with the refutations is that they are using functions where $\lim_{p^m\to\infty}f(p^m)\neq0$ because for some $\epsilon>0$, for any $N$, there will exist some prime $p$ that is bigger than that $N$ such that $f(p)>0$, violating the definition of a limit.
Since $f(p^k)$ approaches $0$, it must have a maximum value, call it $M$. Assume that there are $a$ prime numbers that have a prime power $p^\alpha$ such that $f(p^\alpha)\geq1$. We know that, since $f(p^k)$ approaches $0$, $a$ must be finite. Note that if $p$ is not one of the $a$ such primes, then if $p\not | n$, $f(n)<f(np^\alpha)=f(n)f(p^\alpha).$ Thus, an upper bound for any $f(n)$ is $M^a$.
Let $\epsilon>0$ be given. Let $N$ be such that if $p^\alpha>N$, $f(p^\alpha)<\epsilon/M^a$. Thus, if $n$ is divisible by a prime power greater than $N$ (call it $p^\alpha$), then $f(n)<\epsilon$ since $M^\alpha$ is the upper bound for any value of $f$, so if $n=ap^\alpha$, $f(ap^\alpha)=f(a)f(p^\alpha)\leq(M^a)(\epsilon/M^a)=\epsilon.$
We will now construct a number that must have at least one prime power greater than $N$. Let $p_k$ denote the $k$th prime, and let $\alpha_k$ be the lowest positive integer such that $p_k^{\alpha_k}>N.$ Then, we define
$$N_2=\prod_{\{k:p_k\leq N\}}p_k^{\alpha_k}.$$
By construction, it is impossible for any $n\geq N_2$ to not be dividible by a prime power greater than $N$. Thus, if $n>N_2$, $f(n)<\epsilon.$ Thus, $\lim_{n\to\infty}f(n)=0$.
