How find this$\frac{1}{{{p}_{1}}}+\frac{1}{{{p}_{2}}}+...+\frac{1}{{{p}_{n}}}<10$ Let ${{p}_{1}},{{p}_{2}},...,{{p}_{n}}$ be the prime numbers less than ${{2}^{100}}$.
Prove that
$$\frac{1}{{{p}_{1}}}+\frac{1}{{{p}_{2}}}+...+\frac{1}{{{p}_{n}}}<10$$
This problem is from this Romania National Olympiad 2013,grade 10 -P4  http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3000224&sid=70162fff5664ceb6c25410e6fe0a42f6#p3000224
So  I think this problem have nice methods,it must have  without anlaysis numbers methods
This following is ugly methods By mathlinks

Lemma 1, We have that $$\prod_{p\ prime \leq x}p\leq 4^x$$ for any positive integer $x$.

Proof:

We use induction. The base cases are trivial. Passing from an odd integer to an even one is very easy (as LHS remains the same while the RHS increases). So we only show how to pass from an even integer to an odd one (i.e. prove the inequality for $x=2k+1$). We will use the binomial coefficient $\dbinom{2k+1}{k+1}$. It is easy to see that it is divisible by $\displaystyle \prod_{k+1<p\leq 2k+1}p$ (the product is taken over primes). Therefore $\prod_{p\leq 2k+1}p\leq \left(\prod_{p\leq k+1}p\right)\left(\prod_{k+1<p\leq 2k+1}p\right)\leq 4^{k+1}\dbinom{2k+1}{k+1}\leq 4^{2k+1}$ (we used the induction hypothesis and the fact that $\dbinom{2k+1}{k+1}\leq 2^{2k}$, which is an easy exercise). The lemma is proved.
Lemma 2:(Partial summation) Let $a_n$ ($n\in\mathbb{N}$) be a sequence, so that $a_n=0$ for $n<x_0$, and $S(x)=\displaystyle\sum_{n\leq x} a_n$. Let $f$ be a function with continuous derivative. Then $\sum_{n\leq x} a_nf(n)=S(x)f(x)-\int_{x_0}^{x}S(t)f'(t)\mathrm{d} t$

Proof:Let us note that $$\sum_{n\leq x} a_nf(n)=\sum_{n\leq x} (S(n)-S(n-1))f(n)=S(x)f(x)-$$
 $$\sum_{n\leq x} S(n-1)(f(n)-f(n-1))=S(x)f(x)-\sum_{n\leq x-1} S(n)\int_{n}^{n+1}f'(t)\mathrm{d}t$$.
 As $S$ behaves like a step function constant on $[n,n+1)$, we get that $\sum_{n\leq x}a_nf(n)=S(x)f(x)-\int_{0}^xS(t)f'(t)\ \mathrm{d}t=S(x)f(x)-\int_{x_0}^xS(t)f'(t)\mathrm{d}t$, as $a_n=0$ for $n<x_0$.
With these two lemmas we are ready to prove the problem. From the first lemma we have (by taking logarithms) that $\sum_{p\leq n}\log(p)\leq n\log(4)$, for any integer $n$ so the relation actually holds even if $n$ is any positive real number. We use lemma 2 with $a_n=\log(n)$ if $n$ is prime and $a_n=0$ otherwise. We take $f(x)=\frac{1}{x\log(x)}$. We can take in the lemma $x_0=2$. We have that $$\sum_{p\leq x}\frac{1}{p}=\sum_{n\leq x} a_nf(n)=S(x)f(x)-\int_{2}^{x}S(t)f'(t)\mathrm{d} t=\frac{S(x)}{x\log(x)}+\int_{2}^x\frac{S(t)(1+\log(t))}{t^2\log^2(t)}\mathrm{d}t$$ Using that $S(x)
\leq x\log(4)$ we get that $$\sum_{p\leq x}\frac{1}{p}\leq \frac{\log(4)}{\log(x)}+\int_{2}^x\frac{\log(4)}{t\log^2(t)}\mathrm{d}t+\int_{2}^x\frac{\log(4)}{t\log(t)}\mathrm{d}t$$. The antiderivative of the function in the first integral is $\frac{1}{\log(t)}$, so the first integral is at most $\frac{\log(4)}{\log(2)}=2$, and the antiderivative of the function in the second integral is $\log(\log(x))$. We therefore have that $$\sum_{p\leq x}\frac{1}{p}\leq \frac{\log(4)}{\log(x)}+2+\log(4)(\log(\log(x))-\log(\log(2)))$. For $x=2^{100}$$
we have that $$\sum_{p\leq 2^{100}}\frac{1}{p}\leq \frac{1}{50}+2+\log(4)\log(100)$$, and the last expression is less than $8.405$ (in an olympiad one might use that $e$ is greater than $2.7$ and use this to estimate $\log(2)$ and $\log(10)$)
 A: Denote
$$S=\sum_{i=1}^n\frac{1}{p_i}.$$
Then 
$$S^2=\sum_{1\le i,j\le n}\frac{1}{p_ip_j}< 2\sum_{1\le i\le j\le n}\frac{1}{p_ip_j}$$
and 
$$S^3=\sum_{1\le i,j,k\le n}\frac{1}{p_ip_jp_k}< 6\sum_{1\le i\le j\le k\le n}\frac{1}{p_ip_jp_k}.$$
As a result,
$$\frac{1}{6}S^3+\frac{1}{2}S^2+S<\sum_{k=2}^{p_n^3}\frac{1}{k}<\sum_{k=2}^{2^{300}}\frac{1}{k}<\int_1^{2^{300}}\frac{dx}{x}=300\log 2< 208.$$
It follows that $S<10$.

Edit: The estimate of $S$ could be improved by using the same trick as follows.
For every $m\ge 1$,
$$S^m=\sum_{1\le i_1,\cdots,i_m\le n}\frac{1}{\prod_{j=1}^mp_{i_j}}< m!\sum_{1\le i_1\le\cdots\le i_m\le n}\frac{1}{\prod_{j=1}^mp_{i_j}}.$$
Therefore, for every $m\ge 1$,
$$\sum_{i=1}^m\frac{S^i}{i!}<\sum_{k=2}^{p_n^m}\frac{1}{k}<\sum_{k=2}^{2^{100m}}\frac{1}{k}<\int_1^{2^{100m}}\frac{dx}{x}=100m\cdot\log 2.$$
In particular, letting $m=6$, we can obtain $S<7$.
A: One might want to know how good estimates for $S=\sum_{p\le x}\frac{1}{p}$ one can obtain (as a remark, certainly not suited for a contest problem). Dusard has
explicit estimates in Theorem $6.10$ of the paper arxiv.org/pdf/1002.0442v1.pdf:
$$
\sum_{p\le x}\frac{1}{p}\le B+\log(\log (x))+\frac{1}{10\log(\log(x))}+\frac{4}{15\log(\log(\log(x)))}
$$
for all $x\ge 10372$, and $B\simeq 0.26149 72128 47643.$ For $x=2^{100}$ this yields $\sum_{p\le x}\frac{1}{p}\le 4.708387538$.
