partial sums

**Random Variable** · April 9th 2011, 09:01 PM

Let $\displaystyle \sum_{n=0}^{\infty} a_{n} x^{n}$ be a convergent series.

Let $S_{n}$ be the nth partial sum of the above series.

Then $\displaystyle \sum^{\infty}_{n=1} S_{n} x^{n} = \frac{1}{1-x} \sum^{\infty}_{n=0} a_{n} x^{n}$ .

Should this be obvious? I don't get it.

**Prove It** · April 9th 2011, 09:08 PM

It might help to note that $\displaystyle \frac{1}{1 - x} = \sum_{n = 0}^{\infty}x^n$ ...

**Random Variable** · April 9th 2011, 09:35 PM

$\displaystyle \sum_{n=0}^{\infty} S_{n} x^{n} = \sum^{\infty}_{n=0} \sum^{n}_{k=0} (a_{k}x^{k}) x^{n}$

Reverse the order of summation?

**Random Variable** · April 9th 2011, 11:08 PM

It makes sense when you write out the terms.

$\displaystyle \frac{1}{1-x} \sum^{\infty}_{n=0} a_{n} x^{n} = \frac{a_{0} + a_{1} x + a_{2}x^{2} + ...}{1-x}$

$\displaystyle = (a_{0} + a_{1} x + a_{2}x^{2} + ...)(1 + x + x^{2} + ...) \ , \ \text{for} \ |x| < 1$

$\displystyle= a_{0} + (a_{0}+ a_{1}) x + (a_{0}+a_{1}+a_{2}) x^{2} + ...$

$\displaystyle = \sum^{\infty}_{n=0} S_{n} x^{n}$

**Drexel28** · April 9th 2011, 11:16 PM

Originally Posted by Random Variable

Let $\displaystyle \sum_{n=0}^{\infty} a_{n} x^{n}$ be a convergent series.

Let $S_{n}$ be the nth partial sum of the above series.

Then $\displaystyle \sum^{\infty}_{n=1} S_{n} x^{n} = \frac{1}{1-x} \sum^{\infty}_{n=0} a_{n} x^{n}$ .

Should this be obvious? I don't get it.

I think you're overthinking this, by definition assuming all the proper conditions are met the Cauchy Product says that $\displaystyle \sum_n x^n\sum_n a_n x^n=\sum_n (\sum_{k=0}^n a_k 1)x^n$

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