This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?
Asked
Active
Viewed 296 times
2
-
4It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$. – Fimpellizzeri Nov 23 '18 at 03:26
-
Absolute convergence is a very strong result in analysis. So basically, IF you can show that $\sum_{n=1}^\infty |a_n|$ converges, then you know the sum without absolute value $\sum_{n=1}^\infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier. – Decaf-Math Nov 23 '18 at 03:46
-
Your textbook probably has a typo. – Paramanand Singh Nov 23 '18 at 12:25
4 Answers
2
This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.
Juan123
- 179
1
The series can be written as the sum of positive terms and negative terms.
$$\sum_{i=1}^{\infty}(-3/4)^{i+1} = \sum_{k=0}^{\infty}(3/4)^{2k}-\sum_{k=0}^{\infty}(3/4)^{2k+1} = a_n - b_n$$
Both $a_n$ and $b_n$ being geometric series are convergent.
Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.
Chickenmancer
- 4,347
Ajit Sharma
- 21
1
If $|x| < 1$ then $\displaystyle\sum_{n=0}^{\infty} x^n =\frac1{1-x}$.
Therefore
$\begin{align}\\ \sum_{n=1}^{\infty} \left(\frac{-3}{4}\right)^{n+1} &=\left(\frac{-3}{4}\right)^2\sum_{n=1}^{\infty} \left(\frac{-3}{4}\right)^{n-1}\\ &=\frac{9}{16}\sum_{n=0}^{\infty} \left(\frac{-3}{4}\right)^{n}\\ &=\frac{9}{16}\frac1{1-\left(\frac{-3}{4}\right)}\\ &=\frac{9}{16}\frac{4}{4+3}\\ &=\frac{9}{4\cdot 7}\\ &=\frac{9}{28}\\ \end{align} $
Chickenmancer
- 4,347
marty cohen
- 107,799
