Prove $\sum_{n=0}^\infty \frac{(-1)^n 4}{2n+1}>\frac{9}{5}$ 
Prove $\sum_{n=0}^\infty \frac{(-1)^n 4}{2n+1}>\frac{9}{5}$

So I tried first proving it converges, because otherwise it would be nonsense. I tried with Leibniz since you can pull out the $4$ and it's a decreasing function since $$a_n>a_{n+1}$$ being $a_n=\frac{4}{2n+1}$, it's also always positive since $2n+1>0$ $\forall n\in(0,\infty)$ and $a_n$ tends to $0$ as $n$ approaches infinity, which then it's proven to be a convergent series. But to actually prove that it is bigger than $\frac{9}{5}$ I have little to no ideas, the book suggests to take the first 2nd and 3rd terms of the series but it doesn't make sense to me how to actually do it.
 A: When you have a series $\sum_{n=0}^\infty(-1)^na_n$ with $a_0>a_1>a_2>\cdots$, then the sequence $\left(\sum_{n=0}^{2N}(-1)^na_n\right)_{n\in\Bbb N}$ is decreasing and the sequence $\left(\sum_{n=0}^{2N-1}(-1)^na_n\right)_{n\in\Bbb N}$ is increasing. Furthermore, each number of the form $\sum_{n=0}^{2N}(-1)^na_n$ is greater than any number of the form $\sum_{n=0}^{2N-1}(-1)^na_n$.
In your case, $\sum_{n=0}^1(-1)^na_n=\frac83>\frac95$. It follows from what I wrote above that$$N\geqslant1\implies\sum_{n=0}^N(-1)^na_n>\frac83$$and that therefore $\sum_{n=0}^\infty(-1)^na_n>\frac83\geqslant\frac95.$
A: This sum is equal to $\pi$, which is clearly above $3$ and it (trivially) follows that $\pi>3>\frac{9}{5}$.
Now, it is very easy to see that your series converges to $\pi$, since check the first formula here, from Wikipedia Italia $\sum_{n=0}^\infty \frac{(-1)^n 4}{2n+1} = 4 \cdot \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = 4 \cdot \frac{\pi}{4}$ by Reference 1 above.
A: *

*For series convergence we need: $\displaystyle \lim_n a_n = 0$.


*Since this series converges we can verify that:
$$
\arctan (x) = \sum_{n \ge 0}\dfrac{(-x)^n}{2n + 1}, |x| < 1
$$
