how to find upper and lower bound of sequence

Question

sequence is $$a_x = (-1)^x*sin^2(x)$$ $$ S = \sum\limits_{i=1}^{\infty}(a_i) $$ sum does not converge, but I assume that it should have some not infinite bounds, but I can't understand how to find them

Answer 1

With $a_k=\sin^2k,$ we have to find the exact upper and lower bounds of $\displaystyle s_n=\sum^n_{k=1}(-1)^k\,a_k,$ that's what the question says. It's not very hard to find an explicit form of $s_n$, it's a special version of telescoping sums: if $a_k=b_k+b_{k-1}$, we have $\displaystyle\sum^n_{k=1}(-1)^k\,a_k=(-1)^n\,b_n-b_0.$ Now we have $$\sin^2k=\frac{1-\cos2k}2,$$ and $$\cos2k=\frac{\cos(2k+1)+\cos(2k-1)}{2\,\cos1},$$ so $$b_k=\frac14\,\left(1-\frac{\cos(2k+1)}{\cos1}\right)$$ will turn the trick: $$s_n=(-1)^n\,\frac14\,\left(1-\frac{\cos(2n+1)}{\cos1}\right).$$ So obviously $$s_n\le\frac14\,\left(1+\frac1{\cos1}\right)$$ and $$s_n\ge-\frac14\,\left(1+\frac1{\cos1}\right),$$ and those bounds are exact, since it is (relatively) well known that $\cos(2n+1)$ can be arbitrarily close to $1$ and $-1$ for infinitely many (odd or even) $n$.