Converting partial sum to infinite series

**xEnOn** · April 25th 2011, 05:20 AM

If a partial sum of a series is given this way: $S_{n} = \frac{3^{n+5} + 8n^{2} - 9 }{3^{n}}$

Then, can I say since the $\lim_{n\rightarrow \infty }S_{n} = \frac{3^{n+5} + 8n^{2} - 9 }{3^{n}}=243$ ,

Is this right? Can I do this to find the series of the partial sum equation of

?

Thanks!

**alexmahone** · April 25th 2011, 07:02 AM

Originally Posted by xEnOn

If a partial sum of a series is given this way: $S_{n} = \frac{3^{n+5} + 8n^{2} - 9 }{3^{n}}$

Then, can I say since the $\lim_{n\rightarrow \infty }S_{n} = \frac{3^{n+5} + 8n^{2} - 9 }{3^{n}}=243$ ,

Is this right? Can I do this to find the series of the partial sum equation of

?

Thanks!

No; the series isn't geometric.

Find $T_n$ the traditional way. ( $T_n=S_n-S_{n-1}$ )

**xEnOn** · April 25th 2011, 08:10 AM

But if say I want to find Xn of this:
$\lim_{n\rightarrow \infty }\frac{3^{n+5}+8n^{2}-9}{3^{n}}=\sum_{n=0}^{\infty }X_{n}=243$

Is there a way to solve for $X_{n}$ in the summation?

**alexmahone** · April 25th 2011, 08:16 AM

Originally Posted by xEnOn

But if say I want to find Xn of this:
$\lim_{n\rightarrow \infty }\frac{3^{n+5}+8n^{2}-9}{3^{n}}=\sum_{n=0}^{\infty }X_{n}=243$

Is there a way to solve for $X_{n}$ in the summation?

Use my formula in post #2 to find the nth term.

**xEnOn** · April 25th 2011, 08:37 AM

hmm...I am not very sure how I can use the formula. I haven't tried this before. Is the

?

And then I have something like this:

?
But this looks weird because Tn will become 0.

**alexmahone** · April 25th 2011, 08:42 AM

Originally Posted by xEnOn

hmm...I am not very sure how I can use the formula. I haven't tried this before. Is the

?

And then I have something like this:

?
But this looks weird because Tn will become 0.

The nth term $T_n=S_n-S_{n-1}$

= $\frac{3^{n+5}+8n^2-9}{3^n}-\frac{3^{n+4}+8(n-1)^2-9}{3^{n-1}}$

= $\frac{3^{n+5}+8n^2-9-3^{n+5}-24(n-1)^2+27}{3^n}$

= $\frac{18+8n^2-24(n^2-2n+1)}{3^n}$

= $\frac{-16n^2+48n-6}{3^n}$

**xEnOn** · April 25th 2011, 08:54 AM

But the difference between the sum of (0 to n) and sum of (0 to n-1) is not always the same. So this isn't the kind of usual arithmetic series and I cannot just say this is it $\sum_{n=0}^{\infty }\frac{2(-8n^{2}+24n-3)}{3^{k}}$ right?

And since it is going to be an infinite series, don't I have to take some kind of limits towards infinity?

**alexmahone** · April 25th 2011, 08:59 AM

Originally Posted by xEnOn

But the difference between the sum of (0 to n) and sum of (0 to n-1) is not always the same. So this isn't the kind of usual arithmetic series and I cannot just say this is it $\sum_{n=0}^{\infty }\frac{2(-8n^{2}+24n-3)}{3^{k}}$ right?

And since it is going to be an infinite series, don't I have to take some kind of limits towards infinity?

This series is neither arithmetic nor geometric.

$S_n=\sum_{n=1}^{n}\frac{2(-8n^{2}+24n-3)}{3^{n}}$

**xEnOn** · April 25th 2011, 09:06 AM

ohh.. So it could just be like a summation of the difference between the previous and current term to get the eventual sum of the equation. But would this be very accurate compared to the actual equation?

Thank you so much!!

**alexmahone** · April 25th 2011, 09:10 AM

Originally Posted by xEnOn

ohh.. So it could just be like a summation of the difference between the previous and current term to get the eventual sum of the equation. But would this be very accurate compared to the actual equation?

Thank you so much!!

More than accurate...it is exact!

**xEnOn** · April 25th 2011, 09:14 AM

great! thanks for all the help!

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