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$$S_n=8-n4^{-n}$$

I know that this series is convergent and: $$\sum _{n=1}^{\infty}a_n=\lim_{n\rightarrow\infty}S_n=8$$

But how do I find $a_n$ on its own?

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To get $a_n$ from $S_n$, use $a_n = S_{n} - S_{n-1}$.

In your case, assuming $S_n = 8 - \frac{4}{n}$:

$$a_n = \frac{4}{n-1} - \frac{4}{n} = \frac{4}{n (n-1)}$$

EDIT

When $S_n = 8 - \frac{n}{4^n}$:

$$a_n = \frac{n-1}{4^{n-1}} - \frac{n}{4^{n}} = \frac{4n-4}{4^{n}} - \frac{n}{4^{n}} = \frac{3n-4}{4^{n}}$$

Note that this does not include a constant of summation, say, $a_0$, that would provide the limiting value of $8$.

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  • $\begingroup$ @Babak: I think your comment should have gone with the question, not this answer. $\endgroup$ – Ron Gordon Jan 25 '13 at 4:20
  • $\begingroup$ I apologize, the exponent should be $-n$, not $-1$ All I really needed was that formula, so thanks. $\endgroup$ – Chris A Jan 25 '13 at 4:24

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