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Hint: $$ \frac{n^3}{n!}=\frac{n^2}{(n-1)!}=\frac{1}{(n-1)!}+\frac{n+1}{(n-2)!}=\frac{1}{(n-1)!}+\frac{3}{(n-2)!}+\frac{1}{(n-3)!} $$


This approach is perfectly valid. When we have a series $$\sum_{n=0}^\infty a_nx^n$$ then replacing $x\mapsto x^2$ we get $$\sum_{n=0}^\infty a_nx^{2n}=\sum_{n=0}^\infty b_nx^n$$ which is a power series too with $$b_n=\begin{cases} a_{n/2},&\text{when $n$ is even}\\ 0,&\text{otherwise} \end{cases}$$ Moreover the radius of convergence is ...


You will need $|x|+|y|<1$. That should be necessary and sufficient. This is because, re-arranging, this is $\sum_{n=0}^{\infty} (|x|+|y|)^n$, where $n=k+l$.


HINT:Your method is correct for $\dfrac{|z-w|}{|1-w|}<1$ . For the series $f(z)=\frac{-1}{z}\frac{1}{1-\frac{1}{z}}=\frac{-1}{z}\sum\frac{1}{z^n}$ note that it is valid in the anulus $|z|>1$.


Based on the context of learning, if the textbook said you can differentiate a power series term by term inside circle of the convergence, then just differentiate it. You can prove $c_1 =0 $ and do it recursively. Otherwise, you need to prove you can differentiate a converged power series term by term and still have a converged power series.


We have $|b_n|\to 0$ (Baby Rudin 3.23). Thus, for $|x|<1$, $\sum |b_n||x|^n$ is convergent since $\sum |x|^n$ is (comparison test). The radius of convergence is therefore $\ge 1$. If the radius $R$ were $> 1$, then we should have absolute convergence for $|x|<R$ (follows from proof of Baby Rudin 3.39). This contradicts the non-absolute ...


The relative error for $\pi^k$ after summing $n$ terms is $\approx n^{-k}$. Computing the $k$th root then does little change (the relative error becomes $\frac 1kn^{-k}$). Hence the number of correct digits is essentially $k\log_{10}n$. For any fixed $k$, this does not grow very well if we compare it to what the Borweins managed (in the linked Wikipedia ...


Hint: For $|s|<1$, $$\frac{1}{1-s}=\sum_{k=0}^\infty s^k$$ To get each term in your expression into this form is just a matter of rewriting. Take this simple example: $$\frac{1}{2-x}=\frac{1}{2}\times\frac{1}{1-\frac{x}{2}}=\frac{1}{2}\sum_{k=0}^\infty \left(\frac{x}{2}\right)^k$$

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