# Thread: Finding the nth partial sum and taking the limit?

1. ## Finding the nth partial sum and taking the limit?

I need to find the sum of each series by computing their n-th partial sum and then taking the limit as n -> infinity.

1) Σ (infinity being on top, n=0 on bottom) 5/2^n - 4/3^n

I already took the limit by separating the two, and ending up with:

5/(1-(1/2)) - 4(1-(1/3)) = 4

I just need the n-th partial sum.

2) Σ (infinity being on top, n=1 on bottom) 1/(n^2-(1/4))

I really don't know how to do this one.

Thanks!

2. First hint : $\sum_{j=0}^nx^j = \frac{1-x^{n+1}}{1-x}$

Second hint : $\frac{1}{n^2-1/4}=\frac{1}{n-1/2}-\frac{1}{n+1/2}$

3. Originally Posted by lindz
I need to find the sum of each series by computing their n-th partial sum and then taking the limit as n -> infinity.

1) Σ (infinity being on top, n=0 on bottom) 5/2^n - 4/3^n

I already took the limit by separating the two, and ending up with:

5/(1-(1/2)) - 4(1-(1/3)) = 4

I just need the n-th partial sum.

2) Σ (infinity being on top, n=1 on bottom) 1/(n^2-(1/4))

I really don't know how to do this one.

Thanks!
$\sum_{\ell=0}^n\omega^\ell=\frac{1-\omega^{n+1}}{1-\omega}$. Does that help?