Inequality Of The Sum Of A Series

[anemone]

Prove \(\displaystyle \frac{10}{\sqrt{11^{11}}}+\frac{11}{\sqrt{12^{12}}}+\cdots+\frac{2015}{\sqrt{2016^{2016}}}\gt \frac{1}{10!}-\frac{1}{2016!}\)

 

[Euge]

Hi anemone,

Here is my solution.

Spoiler

Let $n\in \{11,\ldots, 2016\}$. Then $n!^2$ is the product of $k(n+1-k)$ as $k$ ranges from $1$ to $n$. Consider the inequality $$xy\ge x + y - 1 \qquad (x,y\in \Bbb N)$$ which follows from the inequality $(x - 1)(y - 1) \ge 0$. Since equality holds above if and only if $x = 1$ or $y = 1$, we find that $k(n+1-k) > k + (n+1-k) - 1 = n$ for $1 < k < n$. Therefore $n!^2 > n^n$, or $\sqrt{n^n} < n!$. I now estimate

$$\frac{10}{\sqrt{11^{11}}} + \frac{11}{\sqrt{12^{12}}} + \cdots + \frac{2015}{\sqrt{2016^{2016}}}$$
$$=\sum_{n = 11}^{2016} \frac{n-1}{\sqrt{n^n}} > \sum_{n = 11}^{2016} \frac{n-1}{n!} = \sum_{n = 1}^{2016} \left(\frac{1}{(n-1)!} - \frac{1}{n!}\right),$$

which telescopes to $$\frac{1}{10!} - \frac{1}{2016!}.$$

 

[anemone]

Hi anemone,

Here is my solution.

Spoiler

Let $n\in \{11,\ldots, 2016\}$. Then $n!^2$ is the product of $k(n+1-k)$ as $k$ ranges from $1$ to $n$. Consider the inequality $$xy\ge x + y - 1 \qquad (x,y\in \Bbb N)$$ which follows from the inequality $(x - 1)(y - 1) \ge 0$. Since the above equality holds if and only if $x = 1$ or $y = 1$, we find that $k(n+1-k) > k + (n+1-k) - 1 = n$ for $1 < k < n$. Therefore $n!^2 > n^n$, or $\sqrt{n^n} < n!$. I now estimate

$$\frac{10}{\sqrt{11^{11}}} + \frac{11}{\sqrt{12^{12}}} + \cdots + \frac{2015}{\sqrt{2016^{2016}}}$$
$$=\sum_{n = 11}^{2016} \frac{n-1}{\sqrt{n^n}} > \sum_{n = 11}^{2016} \frac{n-1}{n!} = \sum_{n = 1}^{2016} \left(\frac{1}{(n-1)!} - \frac{1}{n!}\right),$$

which telescopes to $$\frac{1}{10!} - \frac{1}{2016!}.$$

Spectacular, Euge! And thanks for participating!(Cool)

 

[Albert1]

Hi anemone,

Here is my solution.

Spoiler

Let $n\in \{11,\ldots, 2016\}$. Then $n!^2$ is the product of $k(n+1-k)$ as $k$ ranges from $1$ to $n$. Consider the inequality $$xy\ge x + y - 1 \qquad (x,y\in \Bbb N)$$ which follows from the inequality $(x - 1)(y - 1) \ge 0$. Since the above equality holds if and only if $x = 1$ or $y = 1$, we find that $k(n+1-k) > k + (n+1-k) - 1 = n$ for $1 < k < n$. Therefore $n!^2 > n^n$, or $\sqrt{n^n} < n!$. I now estimate

$$\frac{10}{\sqrt{11^{11}}} + \frac{11}{\sqrt{12^{12}}} + \cdots + \frac{2015}{\sqrt{2016^{2016}}}$$
$$=\sum_{n = 11}^{2016} \frac{n-1}{\sqrt{n^n}} > \sum_{n = 11}^{2016} \frac{n-1}{n!} = \sum_{n = 1}^{2016} \left(\frac{1}{(n-1)!} - \frac{1}{n!}\right),$$

which telescopes to $$\frac{1}{10!} - \frac{1}{2016!}.$$

marvelous !