Question

Determine if the series converge or diverge. If they converge, what value do they converge to?
-Sum of {((-1)^n *2^(3n-6))/5^(2n+1)} with the series going from n=-2 to infinity.
- Sum of {7/(n+1)(n+4)} with the series going from n=1 to infinity.

On the first problem, I did the root test and found that L=-8/25 which is less than 1 meaning the series converged. I just don't know how to find the value to which it converges to.
For the second problem, I have gotten as far as the Lim as n->Infinity ((7/3)/n(1+(1/n)) + ((-7/3)/n(1+(4/n)). I don't really know where to go from here.

Answer 1

Usually one hint if you have to determine the sum of a series, it means that you usually have a telescoping series or a geometric series (or perhaps some other special series).
\[\begin{align} \sum_{n=-2}^{\infty} \frac{(-1)^n 2^{3n-6}} {5^{2n+1}} &=\sum_{n=-2}^{\infty}\frac{(-1)^n2^{3n}2^{-6}}{5^{2n}5^1}\\
&=\frac{2^{-6}}{5} \sum_{n=-2}^{\infty}\frac{(-1)^n (2^3)^n}{(5^2)^n}\end{align}\]

Do you see how you can combine the terms using the law of exponents? \(a^n * b^n = (ab)^n\)
And then you can re-index your sum so it starts at 0, and check if your number \((a)^n\) if \(|a|<1\)

Answer 2

re-index? i don't really understand what that means.

Answer 3

TO verify your answer go to http://wolframalpha.com
copy and paste the following in the input box
Sum [((-1)^n *2^(3n-6))/5^(2n+1) , {n,-2,Infinity}]
and hit the equal sign

Answer 4

Well your summation starts at \(n=-2\), and the geometric series formula works for when you start at \(n=0\)

Answer 5

(well .. that's for \[ \sum_{n=0}^{\infty}r^n = \frac{1}{1-r}, \text{if } |r|<1\]

Answer 6

So the way I had orignially got to my answer didnt work and was wrong?

Answer 7

well your method just checked for convergence. Since you checked that, you need to find a way to find a value for your sum. And, usually if you need to find a sum , it tends to be telescoping series or a geometric series . In this case, it turns out to be a geometric series

Answer 8

(actually.. if you could identify you had a geometric series right away with \(|r|<1\) (well, with some algebraic manipulation), you wouldn't need to do a convergence test since you know the geometric series does indeed converge for \(|r|<1\), but I guess "observing" this isn't so obvious when you are first learning about it

Answer 9

so the formula you provided a couple comments up is the one for the geometric series? I really have trouble with applying what I learn or see to what I'm supposed to be doing in these math problems.
My issue right now is how I can get from where you left off, which is where I am at right now, to the answer...

Answer 10

Yes the formula I wrote was for the geometric series.
Now at this step:
\[ \frac{2^{-6}}{5} \sum_{n=-2}^{\infty}\frac{(-1)^n (2^3)^n}{(5^2)^n}\]
Using the exponent rule \(a^n \cdot b^n = (ab)^n\)
We get:
\[ \frac{2^{-6}}{5} \sum_{n=-2}^{\infty}\left(\frac{-8}{25}\right)^n \]
Now the formula for the geometric series I wrote above works when the index at the bottom starts at \(n=0\), but you have it at \(n=-2\), so to re-index it to 0, you can let: \(j = n+2\),
so if \(n=-2\), then \(j=-2+2=0\), so the sum starts at \(j=0\).
The upper index is \(\infty\), so adding/subtracting a finite constant wont change it.

Also, it implies that \(n=j-2\), so
\[ \frac{2^{-6}}{5} \sum_{n=-2}^{\infty}\left(\frac{-8}{25}\right)^n =\frac{2^{-6}}{5}\sum_{j=0}^{\infty}\left( \frac{-8}{25}\right)^{j-2}=\frac{2^{-6}}{5}\sum_{j=0}^{\infty} \left( \frac{-8}{25}\right)^{-2} \left( \frac{-8}{25}\right)^j\]

However, I I just don't like the method of introducing the new dummy variable, A slightly faster approach is to use:
\[ \sum_{n=a}^bf(n)=\sum_{n=a+c}^{b+c}=f(n-c)\] It's really the same procedure though, but you don't have to introduce a new dummy variable.