<p>i hope bc isnt gonna be killer hard</p>
<p>boredpirate:</p>
<p>its almost always on the series question...Has your teacher seen the last 5+ years of FR?</p>
<p>i'm screwed for convergence</p>
<p>I have a question I need help with!</p>
<p>If n is a positive integer, then limit as n approaches positive infinity of </p>
<p>1/n * [sin(pi/n)+sin(2<em>pi/n)+. . . + sin (n</em>pi/n)] is?</p>
<p>The answer is 2*pi, but I have no idea how to get that.</p>
<p>Can you give some more terms in the middle of that series? Does it just start from 1 and go to n? If so, then why did you write sin(pi/n) rather than sin(pi)?</p>
<p>EDIT: Also lim n->infinity (1/n)=0 and since the limit of products is the product of the individual limits, the answer should be zero since its 0* (whatever the lim n->infinity of [sin(pi/n)+sin(2<em>pi/n)+. . . + sin (n</em>pi/n)] is).</p>
<p>It goes from 1 to infinity (within the parantheses, it's not restricted on the outside term), I believe. The way I wrote it is exactly what's in the book -- it's slightly vague. </p>
<p>And a correction on the answer -- it's 2/pi. Sorry!</p>
<p>lim n-> inf E (i=0 to n) sin(i*pi/n) (1/n)</p>
<p>which is you taking an n number of rectangles of width 1/n and height sin(i*pi/n) and adding their areas. As n goes to infinity. A bunch of infinitely thin rectangles.</p>
<p>which is the</p>
<p>integral from 0 to pi of sin(x)</p>
<p>which is -cos x] pi 0 = 1 + 1 = 2</p>
<p>They made the mistake of saying that it was the average of the function sin(x) from 0 to pi. They would need another 1/n for that.</p>
<p>That would then be 2/interval = 2/(pi-0) = 2/pi</p>
<p>Hope that answers your question.</p>