You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Finding a formula for a given sequence?
- Thread starter ZeroXz
- Start date
It is easy to see that they are reduced by powers of 3. A geometric series with first term 999 and common ratio 1/3
\(\displaystyle 999\sum_{k=0}^{\infty}\frac{1}{3^{n}}\)
Now, what does it converge to?.
\(\displaystyle \frac{999}{1-\frac{1}{3}}=\frac{2997}{2}=1498.5\)
The closed form sum for an infinite geometric series is \(\displaystyle \frac{a_{1}}{1-r}\)
where \(\displaystyle a_{1}\) is the first term and r is the common ratio.
\(\displaystyle 999\sum_{k=0}^{\infty}\frac{1}{3^{n}}\)
Now, what does it converge to?.
\(\displaystyle \frac{999}{1-\frac{1}{3}}=\frac{2997}{2}=1498.5\)
The closed form sum for an infinite geometric series is \(\displaystyle \frac{a_{1}}{1-r}\)
where \(\displaystyle a_{1}\) is the first term and r is the common ratio.
galactus said:It is easy to see that they are reduced by powers of 3. A geometric series with first term 999 and common ratio 1/3
\(\displaystyle 999\sum_{k=0}^{\infty}\frac{1}{3^{n}}\)
Now, what does it converge to?.
\(\displaystyle \frac{999}{1-\frac{1}{3}}=\frac{2997}{2}=1498.5\)
The closed form sum for an infinite geometric series is \(\displaystyle \frac{a_{1}}{1-r}\)
where \(\displaystyle a_{1}\) is the first term and r is the common ratio.
Omg. Why didn't I see that?
Initially, I noticed that it goes by division of 3. So I got, (1/3^k) * 111, where k starts from -2. But using this formula I can't get the right answer.
Is it because a sequence, k cannot start from negative number ????
Thanks!!
galactus said:Yes, if k started at -2 in \(\displaystyle \frac{1}{3^{k}}\), then you would have \(\displaystyle 3^{2}\cdot 111\), which would diverge.
By the way, this is an infinite series, not a sequence.
How to know \(\displaystyle 3^{2}\cdot 111\) is diverging instead of converging ???
It's rather obvious, isn't it?. It will keep getting larger and larger and larger instead of converging to some finite value.
\(\displaystyle 3^{2}\cdot 111=999\)
\(\displaystyle 3^{3}\cdot 111=2997\)
.
.
.
\(\displaystyle 3^{20}\cdot 111=387033068511\)
How would this ever converge to some finite value?. The larger k gets, the larger the value becomes. So, it diverges.
\(\displaystyle 3^{2}\cdot 111=999\)
\(\displaystyle 3^{3}\cdot 111=2997\)
.
.
.
\(\displaystyle 3^{20}\cdot 111=387033068511\)
How would this ever converge to some finite value?. The larger k gets, the larger the value becomes. So, it diverges.