Wikipedia:Reference desk/Archives/Mathematics/2015 November 28

= November 28 =

Limit of Infinite Sum and Difference
Given $$d(x)=\sin x$$ (where x is a natural number) and $$D(x)=\sum _{m=1}^x\left(d(m)\right)$$, what is the value of $$ \lim_{x \to \infty} \frac{1}{x-1}\sum _{n=1}^x\left(d(n)-\frac{D(n)}{n}\right)^2$$? This is not for a homework assignment or anything, just asking out of pure curiosity. Thanks for your help. 70.54.112.243 (talk) 05:35, 28 November 2015 (UTC)
 * This isn't exactly rigorous but there may be potential to make it so. What you're basically asking for is the average value of (d(n)-D(n)/n)2. Since d oscillates between -1 and 1 with no preference for positive or negative, its average value, namely D(n)/n, converges to 0. So the D(n)/n term can be ignored and the problem reduces to finding the average value of (d(n))2 = sin2 n. But sin2 n = 1/2 - 1/2 cos 2n. Arguing as before, the average value of cos 2n is 0 so the average we want is 1/2. --RDBury (talk) 18:11, 28 November 2015 (UTC)
 * Great answer, thank you! 70.54.112.243 (talk) 22:36, 28 November 2015 (UTC)