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.

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- Oct 18th 2012, 05:17 PMchathlimit of sequences and series-real analysis
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. - Oct 18th 2012, 05:38 PMchathRe: limit of sequences and series-real analysis
using the alternating series test,I have found that the Leibnitz series converges to zero.but is it sufficient for us to say that the sin of that series converges to zero???

- Oct 18th 2012, 07:58 PMTheEmptySetRe: limit of sequences and series-real analysis
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 $\displaystyle \tan^{-1}(1)=\frac{\pi}{4}$

$\displaystyle \sin \left(\lim_{k \to \infty} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} \right)=\sin\left( \frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$