Is the series indicated absolutely convergent, conditionally convergent, or divergent? #rarr\4-1+1/4-1/16+1/64...#
Use the appropriate test.
I get the feeling this is an Alternating Series, but I'm not sure if it's #(-1)^n# or #(-1)^(n+1)# or #(-1)^(n-1)# ??
The exponents seem to be decreasing, i.e. #a^1# , #a^0# , #a^-1# , #a^-2# ...
Use the appropriate test.
I get the feeling this is an Alternating Series, but I'm not sure if it's
The exponents seem to be decreasing, i.e.
3 Answers
It converges absolutely.
Explanation:
Use the test for absolute convergence. If we take the absolute value of the terms we get the series
#4 + 1 + 1/4 + 1/16 + ...#
This is a geometric series of common ratio
Hopefully this helps!
Explanation:
The geometric series converges absolutely, with
Explanation:
This series is definitely an alternating series; however, it also looks geometric.
If we can determine the common ratio shared by all terms, the series will be in the form
Where
We'll need to find the summation using the above format.
Divide each term by the term before it to determine the common ratio
Thus, this series is geometric, with the common ratio
We can write the series as
Recall that a geometric series
Here,
Now, let's determine if it converges absolutely.
Strip out the alternating negative term:
Take the absolute value, causing the alternating negative term to vanish:
Thus,
We see
The series converges absolutely, with