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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

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  • $\begingroup$ 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. $\endgroup$
    – user436658
    May 7, 2020 at 15:19
  • $\begingroup$ @Professor Vector I need to prove that they are finite $\endgroup$
    – IvanIvan
    May 7, 2020 at 15:47

1 Answer 1

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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) $n$.

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