Convergence prob: sum [j=1, infty] [(2/3)^j + (3/4)^j]

[mooshupork34]

I was confused as to how to do this, so any help would be appreciated

For the following series, I have to prove either (1) That it converges, (2) That it diverges, or (3) Whether it diverges or converges cannot be determined.

$\displaystyle \sum_{j=1}^{\infty} (\frac{2}{3})^j + (\frac{3}{4})^j$

 

[galactus]

Use a geometric series:

\(\displaystyle \L\\\displaystyle\sum_{j=1}^{\infty}\left(\frac{2}{3}\right)^{j}=\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+.......\)

\(\displaystyle \L\\\displaystyle\sum_{j=1}^{\infty}\left(\frac{3}{4}\right)^{j}=\frac{3}{4}+\frac{9}{16}+\frac{27}{64}+............\)

Add them. What is the sum?.

 

[mooshupork34]

Well if you use that formula C/1-a for both of them, you get

1/(1-(2/3)) = 3 for the first one
and 1/(1-(3/4)) = 4 for the second one.

So when you add them together, you get 7.

So I'm guessing it converges because each of the individual expressions converges?

 

[galactus]

No, not quite. You're correct about the convergence, but have the wrong sum.

Here's the first:

Factor out 2/3

\(\displaystyle \L\\\frac{2}{3}(\underbrace{1+\frac{2}{3}+\frac{4}{9}+............}_{\text{this is 1/(1-2/3)}})\)

So, you have:

\(\displaystyle \L\\\frac{\frac{2}{3}}{1-\frac{2}{3}}\)

 

[mooshupork34]

No, not quite. You're correct about the convergence, but have the wrong sum.

Here's the first:

Factor out 2/3

\(\displaystyle \L\\\frac{2}{3}(\underbrace{1+\frac{2}{3}+\frac{4}{9}+............}_{\text{this is 1/(1-2/3)}})\)

So, you have:

\(\displaystyle \L\\\frac{\frac{2}{3}}{1-\frac{2}{3}}\)

Ah! I see! Thanks!