一道奇怪的求和题
函数:
\[f(n)=\frac{n^2}{1+n^2}
\]
数列:
\[\sum^{n}_{n=1} m_n = f(1) + f(2) + f(\frac{1}{2}) + ... + f(n) + f(\frac{1}{n})
\]
推导规律:
\[f(1) = \frac{1}{2}\\
f(2) = \frac{2^2}{1+2^2} = \frac{4}{5}\\
f(\frac{1}{2}) = \frac{1}{5}\\
\]
可得:\(f(n) + f(\frac{1}{n}) = 1\)
由此我们可以知道,在数列中存在\(n-1\)对的\(f(n) + f(\frac{1}{n})\),剩下的一项是\(f(1)=\frac{1}{2}\),易得:
\[f(1) + f(2) + f(\frac{1}{2}) + ... + f(n) + f(\frac{1}{n})\\
= \frac{1}{2} + n - 1\\
= n - \frac{1}{2}
\]
最后,得到
\[\sum^{n}_{n=1} m_n = n - \frac{1}{2}
\]
不知道对不对……