Solution check: summation inequality proof by induction I'm not sure if what I've done works or if it's proof enough.
(I need to prove that the inequality is true $\forall n \in \mathbb{N}$).
$\sum_{i=n}^{2n} \frac{i}{2^i} \leq n$
$P(1)$ works. I assume the inequality to be true for $n$.
$(\sum_{i=n}^{2n} \frac{i}{2^i}) + 1\leq n + 1$
$(\sum_{i=n}^{2n} \frac{i}{2^i}) + \frac{2n+1}{2^{2n + 1}}\leq (\sum_{i=n}^{2n} \frac{i}{2^i}) + 1\leq n + 1$
I'm stuck.
 A: To be proved: $\displaystyle \sum_{i=n}^{2n} \frac{i}{2^i} \leq n$
Base case: $n=1$, $\displaystyle \sum_{i=1}^{2} \frac{i}{2^i}=1 \leq 1$ which works. 
Assume that for some $n$, $\displaystyle \sum_{i=n}^{2n} \frac{i}{2^i} \leq n$
Consider, for $n+1$, $\displaystyle \sum_{i=n+1}^{2(n+1)} \frac{i}{2^i} \leq n+1$
$\displaystyle \sum_{i=n+1}^{2n+2} \frac{i}{2^i} \leq n+1$
$\displaystyle \sum_{i=n}^{2n} \frac{i}{2^i}+\frac{2n+1}{2^{2n+1}}+\frac{2n+2}{2^{2n+2}}-\frac{n}{2^{n}} \leq n+1$
Consider now the inequality: 
$\frac{2n+1}{2^{2n+1}}+\frac{2n+2}{2^{2n+2}}-\frac{n}{2^{n}}≤1$
Prove this inequality to be true, and in conjunction with the assumption and the base case the inductive proof will be complete.
EDIT:
To be neat and tidy, you might want to, following your original line of thinking, say: 
Assume that for some $n$, $\displaystyle \sum_{i=n}^{2n} \frac{i}{2^i} \leq n$
Then, $\displaystyle \sum_{i=n}^{2n} \frac{i}{2^i} +1 \leq n+1$
But, $\frac{2n+1}{2^{2n+1}}+\frac{2n+2}{2^{2n+2}}-\frac{n}{2^{n}}≤1$
And so, $\displaystyle \sum_{i=n}^{2n} \frac{i}{2^i} +\frac{2n+1}{2^{2n+1}}+\frac{2n+2}{2^{2n+2}}-\frac{n}{2^{n}} \leq n+1$
Therefore, $\displaystyle \sum_{i=n+1}^{2(n+1)} \frac{i}{2^i} \leq n+1$. 
Then, you can make the proper conclusion statement regarding induction (the inequality holds for base case $n=1$, and so holds for...)
