Prove by the method of mathematical induction that $\sum_{r=2}^{n}\frac{2}{(r+3)(r+5)}=\frac{11}{30}-\frac{2n+9}{(n+4)(n+5)}\,.$ Hence find $\sum_{r=4}^{n+4}\frac{2}{r(r+2)}\,.$ Deduce that $\sum_{r=4}^{n+4}\frac{1}{(r+1)^2}<\frac{9}{40}\,.$

1. State the proposition and prove it is true for the base case $n=2$.

Let $P_n$ be the proposition that $\sum_{r=2}^{n}\frac{2}{(r+3)(r+5)}=\frac{11}{30}-\frac{2n+9}{(n+4)(n+5)}$.
For $n=2$,
$\text{LHS} = \sum_{r=2}^{2}\frac{2}{(r+3)(r+5)} = \frac{2}{(5)(7)} = \frac{2}{35}$
$\text{RHS} = \frac{11}{30} - \frac{2(2)+9}{(2+4)(2+5)} = \frac{11}{30} - \frac{13}{42} = \frac{77-65}{210} = \frac{12}{210} = \frac{2}{35}$
Since LHS = RHS, $P_2$ is true.

2. State the inductive hypothesis for $n=k$.

Assume $P_k$ is true for some integer $k \ge 2$, i.e.,
$\sum_{r=2}^{k}\frac{2}{(r+3)(r+5)}=\frac{11}{30}-\frac{2k+9}{(k+4)(k+5)}$

3. Write the sum for $n=k+1$ and substitute the inductive hypothesis.

We want to show that $P_{k+1}$ is true.
For $n=k+1$,
$\sum_{r=2}^{k+1}\frac{2}{(r+3)(r+5)} = \sum_{r=2}^{k}\frac{2}{(r+3)(r+5)} + \frac{2}{(k+1+3)(k+1+5)}$
$= \frac{11}{30}-\frac{2k+9}{(k+4)(k+5)} + \frac{2}{(k+4)(k+6)}$

4. Combine the fractions algebraically over a common denominator.

$= \frac{11}{30} - \left[ \frac{(2k+9)(k+6) - 2(k+5)}{(k+4)(k+5)(k+6)} \right]$
$= \frac{11}{30} - \left[ \frac{(2k^2 + 12k + 9k + 54) - 2k - 10}{(k+4)(k+5)(k+6)} \right]$
$= \frac{11}{30} - \frac{2k^2 + 19k + 44}{(k+4)(k+5)(k+6)}$

5. Factorise the numerator to cancel with the denominator, matching the form for $P_{k+1}$, and write the conclusion.

$= \frac{11}{30} - \frac{(2k+11)(k+4)}{(k+4)(k+5)(k+6)}$
$= \frac{11}{30} - \frac{2k+11}{(k+5)(k+6)}$
$= \frac{11}{30} - \frac{2(k+1)+9}{((k+1)+4)((k+1)+5)}$
Hence, $P_{k+1}$ is true.
Since $P_2$ is true, and $P_k \Rightarrow P_{k+1}$, by Mathematical Induction, $P_n$ is true for all integers $n \ge 2$.

1. Apply an index substitution to match the required sum with the sum from part (i).

Let $k = r-3$. When $r = 4$, $k = 1$. When $r = n+4$, $k = n+1$.
$\sum_{r=4}^{n+4}\frac{2}{r(r+2)} = \sum_{k=1}^{n+1}\frac{2}{(k+3)(k+5)}$

2. Isolate the first term to match the summation limits, then apply the formula from part (i).

$= \frac{2}{(1+3)(1+5)} + \sum_{k=2}^{n+1}\frac{2}{(k+3)(k+5)}$
Using the result from (i) with $n$ replaced by $n+1$:
$= \frac{2}{24} + \left( \frac{11}{30} - \frac{2(n+1)+9}{(n+1+4)(n+1+5)} \right)$

3. Simplify the fractions to obtain the final expression.

$= \frac{1}{12} + \frac{11}{30} - \frac{2n+11}{(n+5)(n+6)}$
$= \frac{5}{60} + \frac{22}{60} - \frac{2n+11}{(n+5)(n+6)}$
$= \frac{27}{60} - \frac{2n+11}{(n+5)(n+6)}$
$= \frac{9}{20} - \frac{2n+11}{(n+5)(n+6)}$

1. Evaluate the sum in part (ii), which is required for the deduction.

From part (ii), we need to evaluate $\sum_{r=4}^{n+4}\frac{2}{r(r+2)}$.
Isolating the $r=4$ term and re-indexing the rest with $k = r-3$:
$\sum_{r=4}^{n+4}\frac{2}{r(r+2)} = \frac{2}{4(6)} + \sum_{k=2}^{n+1}\frac{2}{(k+3)(k+5)}$
Using the result from (i) with $n$ replaced by $n+1$:
$\sum_{r=4}^{n+4}\frac{2}{r(r+2)} = \frac{1}{12} + \left( \frac{11}{30} - \frac{2(n+1)+9}{(n+1+4)(n+1+5)} \right)$
$= \frac{9}{20} - \frac{2n+11}{(n+5)(n+6)}$

2. Establish an inequality relating the general terms of the series in (iii) and (ii).

Notice the relationship between the denominators:
$(r+1)^2 = r^2 + 2r + 1$
$r(r+2) = r^2 + 2r$
Since $(r+1)^2 > r(r+2)$ for all $r \ge 4$, taking the reciprocal gives:
$\frac{1}{(r+1)^2} < \frac{1}{r(r+2)}$

3. Apply the inequality to the sum and substitute the result from part (ii).

Summing both sides from $r=4$ to $n+4$:
$\sum_{r=4}^{n+4}\frac{1}{(r+1)^2} < \sum_{r=4}^{n+4}\frac{1}{r(r+2)}$
$\sum_{r=4}^{n+4}\frac{1}{(r+1)^2} < \frac{1}{2} \sum_{r=4}^{n+4}\frac{2}{r(r+2)}$
Substitute the expression found for part (ii):
$\sum_{r=4}^{n+4}\frac{1}{(r+1)^2} < \frac{1}{2} \left[ \frac{9}{20} - \frac{2n+11}{(n+5)(n+6)} \right]$
$\sum_{r=4}^{n+4}\frac{1}{(r+1)^2} < \frac{9}{40} - \frac{2n+11}{2(n+5)(n+6)}$

4. Conclude the proof by showing the subtracted algebraic fraction is strictly positive.

For $n \ge 1$, we have $2n+11 > 0$, $n+5 > 0$, and $n+6 > 0$.
Thus, $\frac{2n+11}{2(n+5)(n+6)} > 0$, which means:
$\frac{9}{40} - \frac{2n+11}{2(n+5)(n+6)} < \frac{9}{40}$
Therefore,
$\sum_{r=4}^{n+4}\frac{1}{(r+1)^2} < \frac{9}{40}$