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.

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see other post.

To find the value of the limit as n approaches infinity of the summation (∑) from k=1 to n of (3/4)^k, we can start by expressing the summation in closed form.

Let's denote the summation as S(n), then we have:

S(n) = (3/4)^1 + (3/4)^2 + (3/4)^3 + ... + (3/4)^n

We can notice that this is a geometric series with a common ratio of (3/4). The formula to calculate the sum of an infinite geometric series is:

S = a / (1 - r),

where "a" is the first term and "r" is the common ratio.

In our case, the first term (a) is (3/4)^1 = 3/4, and the common ratio (r) is (3/4).

Using the formula for the sum of an infinite geometric series, we can substitute these values:

S = (3/4) / (1 - 3/4)

Simplifying,

S = (3/4) / (1/4)

S = (3/4) * (4/1)

S = 3

Therefore, the limit as n approaches infinity of the given summation is equal to 3.