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Let the power series to be defined as :

$$\sum\limits_{n=0}^\infty a_nx^n$$

How to determine $a_{n}$ in such cases?

$$\begin{eqnarray}&a)&\quad \sum\limits_{n=0}^\infty \frac {n^{n}} {n!} x^{3n}\\ \\ &b)&\quad \sum\limits_{n=0}^\infty \frac {n}{2^{n}(x+2)^{2n}}\end{eqnarray}$$

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  • $\begingroup$ Could you check (b)? It makes little sense at present... $\endgroup$ – Did Dec 9 '12 at 22:13
  • $\begingroup$ @did why do you think so? $\endgroup$ – 0x6B6F77616C74 Dec 10 '12 at 0:44
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    $\begingroup$ For one thing, because one divides by zero. $\endgroup$ – Did Dec 10 '12 at 5:44
  • $\begingroup$ @did fixed :--) $\endgroup$ – 0x6B6F77616C74 Dec 10 '12 at 22:37
  • $\begingroup$ Still in b), are you sure you mean $\frac{n}{2n}$? $\endgroup$ – Did Dec 11 '12 at 6:00
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Just read off the coefficients. For (a), for instance, you have

$$a_n=\begin{cases} \frac{k^k}{k!},&\text{if }n=3k\\\\ 0,&\text{otherwise}\;. \end{cases}$$

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  • $\begingroup$ Your answer is incorrect (despite three upvotes so far). Your formula yields $a_3=9/2$ but $a_3=1$. $\endgroup$ – Did Dec 9 '12 at 22:13
  • $\begingroup$ @did: Aargh. Mindless. Fixed now; thanks. $\endgroup$ – Brian M. Scott Dec 9 '12 at 22:17
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The two problems are so different from each other that it leaves me wondering if they're even supposed to be part of the same exercise, or if they may be mis-transcribed in some fashion. That said, here's one approach to solving (b):

a) Expand out $\displaystyle\frac{1}{(x+2)^{2n}}$ into a power series in $x$ using the Binomial Series : $$\begin{eqnarray}\frac{1}{(x+2)^{2n}} &=& \frac{1}{\left(2(\frac{x}{2}+1)\right)^{2n}} \\ &=&\frac{1}{2^{2n}}\frac{1}{\left(\frac{x}{2}+1\right)^{2n}} \\ &=&\frac{1}{2^{2n}}\sum_{i=0}^\infty\binom{i+2n-1}{i}\left(-\frac{x}{2}\right)^i \\ &=&\frac{1}{2^{2n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i \end{eqnarray}$$

b) Plug this expansion into your original power series: $$\begin{eqnarray}\sum_{n=0}^\infty\frac{n}{2^n\left(x+2\right)^{2n}} &=&\sum_{n=0}^\infty\frac{n}{2^n}\frac{1}{\left(x+2\right)^{2n}}\\ &=&\sum_{n=0}^\infty\frac{n}{2^n}\left(\frac{1}{2^{2n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i\right) \\ &=&\sum_{n=0}^\infty\left(\frac{n}{2^{3n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i\right) \\ \end{eqnarray}$$

c) Exchange the orders of summation so that the sum over $i$ is your outer summation; you'll be left with something that looks like $\displaystyle\sum_{i=0}^\infty\left(\sum_{n=0}^\infty b_{n,i}\right)x^i$ where $b_{n,i}$ is some complicated expression in $n$ and $i$. Your coefficients $a_i$ are then just $\displaystyle\sum_{n=0}^\infty b_{n,i}$.

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