$\displaystyle

\sum_{n = 1}^{\infty} \frac {1}{4^n}

$

I tried rewriting it to 4^-n and then using the geometric series test 1/1-r

Which makes it to -1/3 convergence. I am not sure if its correct or not...

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- Apr 3rd 2010, 08:37 PMxterminal01Determining series convergence or divergence
$\displaystyle

\sum_{n = 1}^{\infty} \frac {1}{4^n}

$

I tried rewriting it to 4^-n and then using the geometric series test 1/1-r

Which makes it to -1/3 convergence. I am not sure if its correct or not...

- Apr 3rd 2010, 08:40 PMdwsmith
This is a geometric series. Rewrite it so the index starts at zero and (1/4)^n

- Apr 3rd 2010, 08:42 PMdwsmith
I obtained a positive 1/3 when I did it.

- Apr 3rd 2010, 08:50 PMxterminal01
- Apr 3rd 2010, 08:55 PMdwsmith
Since you have$\displaystyle 1/4^n$, you can rewrite it as $\displaystyle (1/4)^n$. This is because 1 to any power is still 1.

Now, because your indexing starts at 1 instead of 0, we can subtract 1 and then add 1.

This will make the index start at 0. The adding one happens to the exponent so now we have $\displaystyle (1/4)^{n+1}$. This can be written as $\displaystyle (1/4)^1*(1/4)^n$.

Of course, $\displaystyle (1/4)^1=(1/4)$ so we just move that out side the summation.

Thus, you have a geometric series times 1/4 which is your a in the formula $\displaystyle a/(1-r)$