Prove that:
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} (n+1)}<2$$
IMC 2015 / Round 2 Problem 1
Solution
Let $S$ denote the given sum. Then:
$$\begin{aligned}
S=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}\left ( n+1 \right )}&= \sum_{n=1}^{\infty}\frac{\sqrt{n}}{n(n+1)} \\
&=\sum_{n=1}^{\infty}\left [ \frac{\sqrt{n}}{n}- \frac{\sqrt{n}}{n+1} \right ]\\
&= \frac{\sqrt{1}}{1}+ \sum_{n=1}^{\infty} \left[ \frac{\sqrt{n+1}}{n+1}-\frac{\sqrt{n}}{n+1} \right]\\
&=1+ \sum_{n=1}^{\infty}\frac{1}{\left ( n+1 \right )\left ( \sqrt{n}+\sqrt{n+1} \right )} \\
&\overset{CS}{\leq} 1+ \frac{1}{4}\sum_{n=1}^{\infty}\left [ \frac{1}{\sqrt{n+1}(n+1)} + \frac{1}{\sqrt{n}(n+1)} \right ]\\
&= 1+ \frac{S}{4}+ \frac{1}{4}\sum_{n=2}^{\infty}\frac{1}{n^{3/2}}
\end{aligned}$$
For the last sum we use the estimation of the integral. Hence:
$$\sum_{n=2}^{\infty} \frac{1}{n^{3/2}} <\int_{1}^{\infty} \frac{1}{x^{3/2}}\, {\rm d}x =2$$
and the result follows, since $\displaystyle S< 1+ \frac{S}{4}+ \frac{1}{2}\Leftrightarrow S<2$.
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} (n+1)}<2$$
IMC 2015 / Round 2 Problem 1
Solution
Let $S$ denote the given sum. Then:
$$\begin{aligned}
S=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}\left ( n+1 \right )}&= \sum_{n=1}^{\infty}\frac{\sqrt{n}}{n(n+1)} \\
&=\sum_{n=1}^{\infty}\left [ \frac{\sqrt{n}}{n}- \frac{\sqrt{n}}{n+1} \right ]\\
&= \frac{\sqrt{1}}{1}+ \sum_{n=1}^{\infty} \left[ \frac{\sqrt{n+1}}{n+1}-\frac{\sqrt{n}}{n+1} \right]\\
&=1+ \sum_{n=1}^{\infty}\frac{1}{\left ( n+1 \right )\left ( \sqrt{n}+\sqrt{n+1} \right )} \\
&\overset{CS}{\leq} 1+ \frac{1}{4}\sum_{n=1}^{\infty}\left [ \frac{1}{\sqrt{n+1}(n+1)} + \frac{1}{\sqrt{n}(n+1)} \right ]\\
&= 1+ \frac{S}{4}+ \frac{1}{4}\sum_{n=2}^{\infty}\frac{1}{n^{3/2}}
\end{aligned}$$
For the last sum we use the estimation of the integral. Hence:
$$\sum_{n=2}^{\infty} \frac{1}{n^{3/2}} <\int_{1}^{\infty} \frac{1}{x^{3/2}}\, {\rm d}x =2$$
and the result follows, since $\displaystyle S< 1+ \frac{S}{4}+ \frac{1}{2}\Leftrightarrow S<2$.
One more way to produce the inequality would be as in the link.
ReplyDelete$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}(n+1)} &=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n} \sqrt{n+1}}\cdot \frac{1}{\sqrt{n+1}} \\
&< \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}\sqrt{n+1}}\cdot \frac{2}{\sqrt{n}+\sqrt{n+1}} \\
&= 2 \sum_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}\\
&= 2\sum_{n=1}^{\infty}\left [ \frac{1}{\sqrt{n}}- \frac{1}{\sqrt{n+1}} \right ] =2
\end{aligned}$$
and this completes the exercise.