please help me to find the answer for the following problem immediately...
limit (sin(sigma k from 1 to infinity)(-1)^k+1/2k-1)) as n goes to infinity.
basically find the limit of SIN(Leibniz formula) as k goes to infinity.
please help me to find the answer for the following problem immediately...
limit (sin(sigma k from 1 to infinity)(-1)^k+1/2k-1)) as n goes to infinity.
basically find the limit of SIN(Leibniz formula) as k goes to infinity.
I think this is what you are asking.
$\displaystyle \lim_{k \to \infty} \sin \left( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} \right)$
Since the sine function is continous, the limit can be moved \lim_{k \to \infty} inside the function.
The sum inside it the infinite series for the arctangent function evaluated at 1.
The $\displaystyle \tan^{-1}(1)=\frac{\pi}{4}$
$\displaystyle \sin \left(\lim_{k \to \infty} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} \right)=\sin\left( \frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$