The discussion centers around proving the inequality involving the sum of a series, specifically comparing the sum $$\frac{10}{\sqrt{11^{11}}}+\frac{11}{\sqrt{12^{12}}}+\cdots+\frac{2015}{\sqrt{2016^{2016}}}$$ to the expression $$\frac{1}{10!}-\frac{1}{2016!}$$. The scope includes mathematical reasoning and potentially exploratory approaches to the proof.
The discussion does not indicate any consensus or resolution regarding the proof of the inequality, as multiple solutions are presented without a clear agreement on their validity or correctness.
The posts do not provide detailed mathematical steps or assumptions underlying the proposed solutions, leaving potential gaps in the reasoning presented.
Participants interested in mathematical proofs, series inequalities, and those looking for collaborative problem-solving approaches may find this discussion relevant.
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 !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 $$\frac{1}{10!} - \frac{1}{2016!}.$$