Proof by induction that $\frac1{n+1} + \frac1{n+2} + \ldots + \frac1{2n} \geq \frac7{12}$ The question is prove that for every integer greater than or equal to 2
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \geq \frac{7}{12}$$
So far I have 
Base case let $p(2)$
\begin{align*} 
\frac{1}{2}+ \frac{1}{2+2} + \frac{1}{4} & \geq \frac{7}{12}\\
\frac{5}{6} & \geq \frac{7}{12}
\end{align*}
therefore base case is true.
Let's assume it is a true statement for $p(n+1)$ for all $n$ greater than or equal to $2.$
\begin{align*}
\frac{1}{(n+1)+1} + \frac{1}{(n+1)+2} + \frac{1}{2(n+1)} \geq \frac{7}{12}\\
\frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{2n+2} \geq \frac{7}{12}
\end{align*}
then I wrote that if $n$ is greater than or equal to $2$ is a true statement then the inequality is true. This is a proof by induction and I'm not sure if I concluded properly 
 A: Consider $P(n+1) - P(n)$.  This difference has only three terms, all the middle terms cancel.  And looking at those three terms you can easily show $P(n+1)-P(n)>0$.  That proves that if $P(n)>7/12$ then $P(n+1)$ must also be.
A: The three dots $\ldots$ are an ellipsis, indicating continuation of the preceding trend of terms up to the final term.
There are only two terms in the $n=2$ case. The ellipsis form of the formula really only definitely indicates the start point, the end point and the general nature of interim points; it doesn't set a minimum number of terms or indeed the maximum number of terms - you are left to infer the exact quantity. 
In this case, there are $n$ terms in each sum, one each for all the numbers from $n+1$ to $2n$. So for the sake of clarity, here are a few terms:
$$\begin{align}
\frac{1}{3} + \frac{1}{4} &= \frac{7}{12} \tag{n=2} \\[0.5em]
\frac{1}{4} + \frac{1}{5} + \frac{1}{6} &= \frac{37}{60} \tag{n=3} \\[0.5em]
\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} &= \frac{533}{840} \tag{n=4} \\[0.5em]
\end{align}$$
You can see that that $\frac{1}{3}$ has been replaced by $\frac{1}{5} + \frac{1}{6}$ in going from $n=2$ to $n=3$, and  $\frac{1}{4}$ has been replaced by $\frac{1}{7} + \frac{1}{8}$ in going from $n=3$ to $n=4$. This should indicate the correct line of the induction argument.
A: what would be p(n+1)? think about where that falls in your solution. p(n+1) is 1/(2n+2)
