n
lim(n→+∞) ∑ (3/4)^k =
k=1
the n is supposed to be above the sigma sign and the k=1 is supposed to be below the sigma sign. thanks.
http://en.wikipedia.org/wiki/Geometric_series#Sum
sum=3/4 * 1/(3/4)= 3/4*4/3=1
check that.
could you also tell me what the series will be ?
thanks.
To evaluate the given expression, let's break it down step by step.
First, let's understand the elements of the expression:
- The symbol ∑ represents the summation notation, indicating that we need to calculate the sum of a series.
- The expression (3/4)^k represents a term in the series. Here, k is a variable that takes on different values as the series progresses.
- The limit lim(n→+∞) means that we need to find the value that the summation approaches as n approaches infinity.
Now, let's simplify the expression:
In this case, the variable n above the sigma sign indicates that we need to find the sum of terms up to a certain value of n. However, since n is approaching infinity, we can rewrite the expression without explicitly mentioning n.
If we observe the series, we can see that it follows a geometric progression with a common ratio of (3/4).
So, the series can be written as:
∑ (3/4)^k = (3/4)^1 + (3/4)^2 + (3/4)^3 + ...
This is an infinite geometric series. The formula to find the sum of an infinite geometric series is:
Sum = a / (1 - r)
Where:
a = first term of the series = 3/4
r = common ratio of the series = 3/4
Using the formula, we can calculate the sum:
Sum = (3/4) / (1 - 3/4)
= (3/4) / (1/4)
= 3
Therefore, the limit of the given expression as n approaches infinity is 3.