MHB Inequality Of The Sum Of A Series

anemone
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Prove $$\frac{10}{\sqrt{11^{11}}}+\frac{11}{\sqrt{12^{12}}}+\cdots+\frac{2015}{\sqrt{2016^{2016}}}\gt \frac{1}{10!}-\frac{1}{2016!}$$
 
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Hi anemone,

Here is my solution.

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 $$\frac{1}{10!} - \frac{1}{2016!}.$$
 
Euge said:
Hi anemone,

Here is my solution.

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)
 
Euge said:
Hi anemone,

Here is my solution.

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 !
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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