Problem source

\begin{questionparts}
	 \item Prove that 
\[
\sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4)=\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5)
\]
and deduce that 
\[
\sum_{r=1}^{n}r^{5}<\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5).
\]
\item Prove that, if $n>1,$ 
\[
\sum_{r=0}^{n-1}r^{5}>\tfrac{1}{6}(n-5)(n-4)(n-3)(n-2)(n-1)n.
\]
\item Let $\mathrm{f}$ be an increasing function. If the limits
\[
\lim_{n\rightarrow\infty}\sum_{r=0}^{n-1}\frac{a}{n}\mathrm{f}\left(\frac{ra}{n}\right)\qquad\mbox{ and }\qquad\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\frac{a}{n}\mathrm{f}\left(\frac{ra}{n}\right)
\]
both exist and are equal, the definite integral ${\displaystyle \int_{0}^{a}\mathrm{f}(x)\,\mathrm{d}x}$
is defined to be their common value. Using this definition, prove
that 
\[
\int_{0}^{a}x^{5}\,\mathrm{d}x=\tfrac{1}{6}a^6.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item Claim: \[
\sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4)=\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5)
\]
Proof: (By Induction)

Base case: (n=1)

\begin{align*}
LHS &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5! \\
RHS &= \frac16 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 5!
\end{align*}

Therefore the base case is true.

Inductive step: Suppose our statement is true for some $n=k$, then consider $n = k+1$

\begin{align*}
\sum_{r=1}^{k+1} r(r+1)(r+2)(r+3)(r+4) &= \sum_{r=1}^{k} r(r+1)(r+2)(r+3)(r+4) + (k+1)(k+2)(k+3)(k+4)(k+5) \\
&\underbrace{=}_{\text{assumption}} \frac16 k(k+1)(k+2)(k+3)(k+4)(k+5) +  (k+1)(k+2)(k+3)(k+4)(k+5)  \\
&= (k+1)(k+2)(k+3)(k+4)(k+5) \l \frac{k}{6} +1\r \\
&= \frac16 (k+1)(k+2)(k+3)(k+4)(k+5)(k+6)
\end{align*}

Therefore our statement is true for $n = k+1$.

Therefore since our statement is true for $n=1$ and if it is true for $n=k$ then it is true for $n = k+1$ by the principle of mathematical induction it is true for all $n \geq 1$

Since \begin{align*}
\sum_{r=1}^{n}r^5 &< \sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4) \\
&= \frac16 n(n+1)(n+2)(n+3)(n+4)(n+5) 
\end{align*}

\item 
\begin{align*}\sum_{r=0}^{n-1} r^5 &> \sum_{r=0}^{n-1} (r-4)(r-3)(r-2)(r-1)r \\
&= \sum_{r=0}^{n-5} r(r+1)(r+2)(r+3)(r+4) \\
&= \frac16 (n-5)(n-4)(n-3)(n-2)(n-1)n
\end{align*}
\item Let $f(x) = x^5$ \begin{align*}
S_{1,n} &= \sum_{r=0}^{n-1}\frac{a}{n}f\left(\frac{ra}{n}\right) \\
&=  \sum_{r=0}^{n-1}\frac{a}{n}\left(\frac{ra}{n}\right)^5 \\ 
&=\frac{a^6}{n^6} \sum_{r=0}^{n-1}r^5\\ 
\end{align*} 

Therefore $\frac{a^6}6 \frac{(n-5)(n-4)(n-3)(n-2)(n-1)n}{n^6} < S_{1,n} < \frac{a^6}6  \frac{(n-1)n(n+1)(n+2)(n+3)(n+4)}{n^6}$ and so $\lim_{n\to\infty} S_{1,n} = \frac{a^6}{6}$.

Similarly, 

\begin{align*}
S_{2,n} &= \sum_{r=1}^{n}\frac{a}{n}f\left(\frac{ra}{n}\right) \\
&= \sum_{r=1}^{n}\frac{a}{n}\left(\frac{ra}{n}\right)^5 \\
&= \frac{a^6}{n^6} \sum_{r=1}^{n} r^5
\end{align*}

Therefore $\frac{a^6}6 \frac{(n-4)(n-3)(n-2)(n-1)n(n+1)}{n^6} < S_{2,n} < \frac{a^6}6  \frac{n(n+1)(n+2)(n+3)(n+4)(n+5)}{n^6}$ and so $\lim_{n\to\infty} S_{2,n} = \frac{a^6}{6}$.

Since both limits exist are are equal, we have

\[
\int_{0}^{a}x^{5}\,\mathrm{d}x=\tfrac{1}{6}a^6.
\]

\end{questionparts}