# Thread: Converting partial sum to infinite series

1. ## Converting partial sum to infinite series

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

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

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

Thanks!

2. Originally Posted by xEnOn
If a partial sum of a series is given this way: $S_{n}&space;=&space;\frac{3^{n+5}&space;+&space;8n^{2}&space;-&space;9&space;}{3^{n}}$

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

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. ($T_n=S_n-S_{n-1}$)

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

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

4. Originally Posted by xEnOn
But if say I want to find Xn of this:
$\lim_{n\rightarrow&space;\infty&space;}\frac{3^{n+5}+8n^{2}-9}{3^{n}}=\sum_{n=0}^{\infty&space;}X_{n}=243$

Is there a way to solve for $X_{n}$ in the summation?
Use my formula in post #2 to find the nth term.

5. 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.

6. 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 $T_n=S_n-S_{n-1}$

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

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

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

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

7. 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 $\sum_{n=0}^{\infty&space;}\frac{2(-8n^{2}+24n-3)}{3^{k}}$ right?

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

8. 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 $\sum_{n=0}^{\infty&space;}\frac{2(-8n^{2}+24n-3)}{3^{k}}$ 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.

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

9. 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!!

10. 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!

11. great! thanks for all the help!