# how to find upper and lower bound of sequence

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

• I guess you mean the exact upper/lower bounds of the partial sums (which are finite, indeed), right? It's a bit cumbersome, but doable.
– user436658
May 7, 2020 at 15:19
• @Professor Vector I need to prove that they are finite May 7, 2020 at 15:47

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) $$n$$.