Does the following equation have only 1 solution of $n=2$? Conjecture:
If $$I=\frac{1}{1+p_{n+1}}+\sum_{k=1}^{n}\frac{1}{p_k}$$
(where $p_n$ denotes the $n$'th prime)
then $n=2$ is the only natural number for $n$ that makes $I$ an integer.
All I really understand to do is to input numbers in for $n$. Beyond that, however, I am at a loss.
I attempted using the knowledge that 
$$\sum_{k=1}^n \frac{1}{p_k}$$
Is never an integer (for $n>1$), but that really lead nowhere.
I am looking for a proof of my conjecture.
 A: Bring all the fractions to the same denominator and add them to get :
$$\frac{A}{p_1p_2\ldots p_k(p_{k+1}+1)}$$
Note that in the numerator $A$ the only number that isn't divisible with $p_n$ is $$(p_{n+1}+1)p_1p_2\ldots p_{n-1}$$
But for the expression to be an integer we must have $p_n \mid A$ so :$$p_n \mid p_{n+1}+1$$ but because $p_{n+1}>p_n$ it follows that : $$p_{n+1} \geq 2p_n-1$$
Now use Bertrand's postulate :
There's a prime between $a$ and $2a-2$ for every $a \geq 3$ to see that $p_n=2,3$ are the only solutions and this leads for $n=2$ or $n=3$ but just $n=3$ leads to an integer  .
A: Assume $n\ge 3$.
By Bertrand's postulate, we have $1+p_{n+1}\le 2p_n$. If $1+p_{n+1}<2p_n$ then $p_n$ cannot divide $1+p_{n+1}$. Otherwise, if $1+p_{n+1}=2p_n$, then $p_2=3$ does not divide $1+p_{n+1}$. (This case actually never happens when $n>2$, but it is simpler just to handle it anyway). In either case there is a $p_j$ with $j\le n$ such that $p_j$ doesn't divide $1+p_{n+1}$.
Therefore, if you put everything on the common denominator $(1+p_{n+1})\prod_{k=1}^n p_k$, this denominator contains exactly one factor of $p_j$, and the only term in the numerator of the sum that is not divisible by $p_j$ is the one arising from $1/p_j$; therefore the numerator is not itself divisible by $p_j$, and the sum thus cannot be an integer.
