If 1/n is the sum of first n terms. then what is the product of 2014 terms…

Given the partial sums S_n = 1/n.

The nth term equals the difference of consecutive partial sums:

• For n ≥ 2: t_n = S_n - S_{n-1} = 1/n - 1/(n-1) = -1/[n(n-1)].

• For n = 1: t_1 = S_1 = 1.

Therefore the product of the first 2014 terms is

• P = t_1 · ∏_{n=2}^{2014} t_n = 1 · ∏_{n=2}^{2014} ( -1/[n(n-1)] ) = (-1)^{2013} / ( ∏_{n=2}^{2014} n · ∏_{n=2}^{2014} (n-1) ).

• Evaluate the products: ∏_{n=2}^{2014} n = 2014! and ∏_{n=2}^{2014} (n-1) = 2013!.

So P = (-1)^{2013}/(2014!·2013!) = -1/(2014!·2013!).