Expanding Power Series: (1- $\frac{2n}{(n+1)^2})^2$

[stunner5000pt]

can someone tell mehow to expand $$(1 - \frac{2n}{(n+1)^2})^2$$ into power series

or at least direct me to the formula!

 

[stunner5000pt]

actaully i wnated to investigate $$\lim_{n \rightarrow \infty} (1- \frac{1}{2} (\frac{n-1}{n+1})^2$$ but i used L'Hopital's rule and found that it would be 1/2.

Thanks anyway!

 

[thepatientmental]

To expand (1 - $\frac{2n}{(n+1)^2})^2$ into a power series, we can use the binomial theorem. The formula for the binomial theorem states that for any real number x and any positive integer n, $(1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$. In order to use this formula for our expression, we can rewrite (1 - $\frac{2n}{(n+1)^2})^2$ as (1 - $\frac{2n}{(n+1)^2}) \cdot$ (1 - $\frac{2n}{(n+1)^2})$. Then, we can apply the binomial theorem to each factor separately.

For the first factor, we have x = -$\frac{2n}{(n+1)^2}$ and n = 2. Plugging these values into the formula, we get $(1 - \frac{2n}{(n+1)^2}) = \sum_{k=0}^{2} \binom{2}{k} (-\frac{2n}{(n+1)^2})^k$. Simplifying, we get $1 - \frac{4n}{(n+1)^2} + \frac{4n^2}{(n+1)^4}$.

For the second factor, we have x = -$\frac{2n}{(n+1)^2}$ and n = 2 again. Plugging these values into the formula, we get $(1 - \frac{2n}{(n+1)^2}) = \sum_{k=0}^{2} \binom{2}{k} (-\frac{2n}{(n+1)^2})^k$. Simplifying, we get $1 - \frac{4n}{(n+1)^2} + \frac{4n^2}{(n+1)^4}$.

To get the final expression, we can multiply these two series together using the rules of power series multiplication. This will give us a series with terms of the form $c_kx^k$, where $c_k$ is a constant and x is our variable. Simplifying, we get:

$(1 - \frac{2n}{(n+1)^2})^