# Thread: Sum of n terms and sum to infinity

1. ## Sum of n terms and sum to infinity

The first three terms of a geometric series are 2 , -1/2 and 1/8 respectively . Find the smallest value of n such that the difference between the sum of the first n terms and the sum to infinity is less than $10^{-5}$ .

Just wondering which is greater , Sum to infinity or the sum of the n terms ...

2. Due to the fact that the reason is $q = -\frac{1}{4} < 0$ the sum of the n first terms is sometimes greater sometimes lower than the sum to infinity (one term out of two is positive and the other one negative). You have to consider the absolute value of the difference.

$\sum_{k=0}^{+\infty}2 q^k - \sum_{k=0}^{n}2 q^k = \sum_{k=n+1}^{+\infty}2 q^k = 2 q^{n+1} \sum_{k=n+1}^{+\infty}q^{k-n-1} = 2 q^{n+1} \sum_{j=0}^{+\infty}q^j$

3. Here the first term(a) is 2
common ratio is -1/4
I have given you the formula for infinite terms
condition is

|r| < 1

for n terms the formula is
$
S_n=\frac{a(r^n-1)}{r-1}

$

do it now

4. ## Re:

THis is what i have done ..

$S_{\infty}-S_n=\frac{a}{1-r}-\frac{a(1-r^n)}{1-r}$

$=\frac{ar^n}{1-r}$

$=\frac{2(-\frac{1}{4})^n}{\frac{5}{4}}$

$=1\frac{3}{5}(-\frac{1}{4})^n<0.00005$

$(-\frac{1}{4})^n<3.125\times10^{-5}$ , which is undefined ...

Where have i gone wrong ?

5. $
\left|\frac{8}{5}\left(-\frac{1}{4}\right)^n\right| < 10^{-5}
$

$\left(\frac{1}{4}\right)^n < \frac{5 \cdot 10^{-5}}{8}$

$
\left(\frac{1}{4}\right)^n < 6.25 \cdot 10^{-6}$

n = 9 should do it.

6. ## Re:

Oh i see , we can use modulus in such cases right ? Why is it so ?

7. Because in your question it is asked to find the difference and as was your doubt( which sum is greater ) value can be negative ,we only need to find the positive value ( difference)