# Thread: limit of sequences and series-real analysis

1. ## limit of sequences and series-real analysis

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.

2. ## Re: 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???

3. ## Re: limit of sequences and series-real analysis

Originally Posted by chath

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.

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

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 $\tan^{-1}(1)=\frac{\pi}{4}$

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