if you don't have any fractions, there's no way it will converge
maybe you have some exponents? Use ^ if so
and also parentheses
what you have, as written, is just some multiple of n
Also, if you want to sum from 1 to ∞, maybe something like ∑(1..∞)
In other words, write something that makes sense.
Does the series converge or diverge? If it converges, what is the sum? Show your work.
∑∞n=1−4(−12)n−1
6 answers
the fraction is -1/2
well,
∞
∑ (-1/2)^n = -1/3
n=1
because it's just a geometric series with
a = -1/2
r = -1/2
I know that's not exactly what you have, but it should not be hard to adapt it.
∞
∑ (-1/2)^n = -1/3
n=1
because it's just a geometric series with
a = -1/2
r = -1/2
I know that's not exactly what you have, but it should not be hard to adapt it.
That didn’t help at all. What is the answer with complex a explanation?
A complex*
frist you have to find the frist term and the common ratio by sub
how? well my submitting 1 for n in
-4x(-1/2)^n-1
this will give me a1=-4x(-1/2)1-1
then identify the common ratio
r=-1/2
simplify the expression
a1=-4
r=-1/2
substituting -4nfor a1 and -1/2 for r into this formula
s=a1/1-r
s=-4/1-(-1/2)
simplify it and this gavew me
s=- 8/3
I think these series is converge because of how we trie to simplify them together and we used a formula that gave us a sum.this expression has a limit
how? well my submitting 1 for n in
-4x(-1/2)^n-1
this will give me a1=-4x(-1/2)1-1
then identify the common ratio
r=-1/2
simplify the expression
a1=-4
r=-1/2
substituting -4nfor a1 and -1/2 for r into this formula
s=a1/1-r
s=-4/1-(-1/2)
simplify it and this gavew me
s=- 8/3
I think these series is converge because of how we trie to simplify them together and we used a formula that gave us a sum.this expression has a limit
1 answer