# Thread: Another Sum Of This Series Problem

1. ## Another Sum Of This Series Problem

Now in this problem I've attached, I was almost positive the answer would be 0 since the denominator will be approaching an infinitely large number, but I guess this isn't the case. Could someone tell me why the answer isn't 0 and how to go about doing this problem?

2. Originally Posted by fattydq
Now in this problem I've attached, I was almost positive the answer would be 0 since the denominator will be approaching an infinitely large number, but I guess this isn't the case. Could someone tell me why the answer isn't 0 and how to go about doing this problem?
It's not zero since you're talking about a sum, not a limit. This problem is just a geometric series.

$\displaystyle \sum_{n=0}^{\infty}\left(\frac{1}{\sqrt{31}}\right )^n =$ $\displaystyle \frac{1}{1-\frac{1}{\sqrt{31}}} = -\frac{31}{\sqrt{31}-31}$

3. Hello, fattydq!

$\displaystyle \sum^{\infty}_{n=0}\frac{1}{(\sqrt{31})^n}$

Now in this problem I've attached, I was almost positive the answer would be 0 . no
since the denominator will be approaching an infinitely large number.
Be careful about what you're referring to . . .

The terms are approach 0 . . . this is true.
. . But the answer (what they are asking for) is the sum of the series.

We have: .$\displaystyle \sum^{\infty}_{n=0}\left(\frac{1}{\sqrt{31}}\right )^n \;=\;1 + \frac{1}{\sqrt{31}} + \left(\frac{1}{\sqrt{31}}\right)^2 + \left(\frac{1}{\sqrt{31}}\right)^3 + \hdots$

This is a geometric series with first term $\displaystyle a = 1$ and common ratio $\displaystyle r = \frac{1}{\sqrt{31}}$

Its sum is: .$\displaystyle S \;=\;\frac{1}{1-\frac{1}{\sqrt{31}}} \;=\;\frac{\sqrt{31}}{\sqrt{31}-1} \;=\;\frac{\sqrt{31}(\sqrt{31}+1)}{30}$