# Thread: Mth101 Q2_8

1. ## Mth101 Q2_8

find the sum of series
∑(-3/4)^k-1 for k=1 to ∞

2. Originally Posted by TAHIR
find the sum of series
∑(-3/4)^k-1 for k=1 to ∞
Is this
$\sum_{k = 0}^{\infty} \left ( \frac{3}{4} \right ) ^{k - 1}$

or

$\sum_{k = 0}^{\infty} \left [ \left ( \frac{3}{4} \right ) ^k - 1 \right ]$

-Dan

3. Originally Posted by topsquark
Is this
$\sum_{k = 0}^{\infty} \left ( \frac{3}{4} \right ) ^{k - 1}$

or

$\sum_{k = 0}^{\infty} \left [ \left ( \frac{3}{4} \right ) ^k - 1 \right ]$

-Dan
Am I missing something Dan? Isn't $\sum_{k=0}^{ \infty} \left( \frac {3}{4} \right)^{k - 1} = \sum_{k=0}^{ \infty} \left( \frac {3}{4} \right)^{-1} \left( \frac {3}{4} \right)^k = \left( \frac {4}{3} \right) \sum_{k = 0}^{ \infty} \left( \frac {3}{4} \right)^k$

Forgive me if I said something stupid, I'm not fully awake at the moment

4. Originally Posted by Jhevon
Am I missing something Dan? Isn't $\sum_{k=0}^{ \infty} \left( \frac {3}{4} \right)^{k - 1} = \sum_{k=0}^{ \infty} \left( \frac {3}{4} \right)^{-1} \left( \frac {3}{4} \right)^k = \left( \frac {4}{3} \right) \sum_{k = 0}^{ \infty} \left( \frac {3}{4} \right)^k$

Forgive me if I said something stupid, I'm not fully awake at the moment
You aren't missing anything. I'm just asking for clarification on the specific problem. The method used to solve them will be fairly similar. (Though apparently I missed the "-" on the 3/4.)

-Dan

5. Originally Posted by topsquark
You aren't missing anything. I'm just asking for clarification on the specific problem. The method used to solve them will be fairly similar. (Though apparently I missed the "-" on the 3/4.)

-Dan
Ok. Yeah, I see now. apparently it was too early in the morning for me . Yeah, I saw that I left out the minus too, but whatever

6. Originally Posted by TAHIR
find the sum of series
∑(-3/4)^k-1 for k=1 to ∞
It is a geometric series