Trying to solve a limit without Taylor series [don't put on hold] For instance in my recent post: 
I have this limit to find $$\lim_{n\to \infty }\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots+\left(-1\right)^{n-1}\cdot \frac{1}{2n-1}\right)=\text{ ?}$$
and we know too this integral $$I_n=\int _0^1\:\frac{x^n}{x^2+1}dx$$ and that relation for recurrence: $$I_{2n}\:\cdot \:\left(-1\right)^{n-1}\:=\:\frac{\left(-1\right)^{n-1}}{2n-1}-\:\left(-1\right)^{n-1}\cdot I_{2n-2}$$
Okay and now how we can continue, if we know recurrence relation because my teacher adviced me to use this and I don't know what helps me, because if I put value for n>2, I'll find some terms and I need to find the sums...
okay so tell me someone if that recurrence relation can helps me and if not put on hold...
 A: (Not sure if this is the method your teacher is looking for. Probably not...)
You're looking for the infinite sum:$\renewcommand{\d}[1]{\operatorname{d}\!{#1}}$
$$1-\frac13+\frac15-\frac17+\dotsb$$

Now, note that:
  $$\int_0^1t^{n-1}\d t=\left[\frac{t^n}n\right]_0^1=\frac1n$$
  This is central to how I'm going to solve this problem. In particular, letting $n=0,2,4,$ etc., this gives us the family of equalities: $\int_0^1\d t=1,\int_0^1t^2\d t=\frac13,\int_0^1t^4\d t=\frac15,$ etc.

Thus, we have:
\begin{align}
1-\frac13+\frac15-\frac17+\dotsb&=\int_0^1\d t-\int_0^1t^2\d t+\int_0^1t^4\d t-\dotsb\\
&=\int_0^1(1-t^2+t^4-\dotsb)\d t
\end{align}
You may recognize $1-t^2+t^4-\dotsb$ as a geometric series. (If you forgot: $1+u+u^2+\dotsb=\dfrac1{1-u}$.) Here we have $u=-t^2$, so we have:

$$1-t^2+t^4-\dotsb=\frac1{1+t^2}$$

Continuing, we have:
\begin{align}
1-\frac13+\frac15-\frac17+\dotsb&=\int_0^1(1-t^2+t^4-\dotsb)\d t\\
&=\int_0^1\frac1{1+t^2}\d t\\
&=\left[\arctan t\right]_0^1\\
&=\arctan1-\arctan0\\
&=\frac\pi4-0\\
&=\frac\pi4
\end{align}
A: I'm going to use the fact that the series:
$\displaystyle 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...+\left(-1\right)^{n-1}\cdot \frac{1}{2n-1}$ 

is actually taylor series expansion of $tan^{-1}(x)$ at $x= 1$.

The function $tan^{-1}(x)$ satisfies the differential equation

$\displaystyle (x^{2}+1)\frac{d^2y(x)}{dx^2}+ 2x\frac{dy(x)}{dx} = 0$

with $y(0)=0, y^{'}(0)=1$. Finding a recurrence for taylor coefficients $u(n)$ of $y(x)$ must solve your case now.

From initial conditions of differential equations, we can say $u(0) = 0, u(1) = 1$.
Replacing $y(x)$ by $\displaystyle\sum_{n=0}^\infty u(n) x^{n}$ in the differential equation, we get 
$\displaystyle2x\sum_{n=1}^\infty nu(n) x^{n-1} + (x^2+1)\sum_{n=2}^\infty n(n-1)u(n) x^{n-2} = 0$.
By collecting powers of $x^n$ and shifting summation indices, 
$\displaystyle\sum_{n=0}^\infty (2n + n(n-1))u(n) x^{n}+ \sum_{n=0}^\infty (n+1)(n+2)u(n+2) x^{n} = 0$

By comparing coefficients of $x^n$, we get this recurrence relation for $n \ge 0$:
$\displaystyle (n^2 + n)u(n) + (n^2 + 3n + 2)u(n+2) = 0 $
Cancelling $(n+1)$ we get, $nu(n) + (n+2)u(n+2) = 0$ .

So the recurrence relation must be $nu(n) + (n+2)u(n+2) = 0$ with $u(0) = 0, u(1) = 1$.
