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A clever solution to find the expected value of a geometric r.v. is those employed in this video lecture of the MITx course "Introduction to Probability: Part 1 - The Fundamentals" (by the way, an extremely enjoyable course) and based on (a) the memoryless property of the geometric r.v. and (b) the total expectation theorem.
For r = 0 r = 0 the limit is obvious 0 0. Now let r ∈ (0, 1) r ∈ (0, 1), we can write r = r = with p ∈ (1, ∞) p ∈ (1, ∞), So. For r = 1 r = 1 the sequence converges to 1 1 as it is always 1 1. Now we have to check r <0 r <0 Which is a alternating sequence, so it converges when (−r n (− r n converges to zero. So all in all we see:
1. For any geometric series, if |r| <1 | r | <1, then your series will converge. Your reasoning is perfectly sound. If a series converges, then the limit of its corresponding sequence is zero. However the question was asking about the sequence. It didn't state that it was a series.
Since the sequence is geometric with ratio r, a2 = ra1,a3 = ra2 = r2a1, and so on. With this fact, you can conclude a relation between a4 and a1 in terms of those two and r. With the former two known, you can solve for r. From there, the formula for the sum of the first n terms of a geometric sequence, with ratio r and first term a1, is given by.
When I look on Wolfram Alpha it says that the partial sum formula for ∑ni = 1i ⋅ xi is: n ∑ i = 1i ⋅ xi = (nx − n − 1)xn + 1 + x (1 − x)2. On this question, an answer said that the general formula for the sum of a finite geometric series is: n − 1 ∑ k = 0xk = 1 − xn 1 − x. But if I substitute my (i ⋅ xi) into the formula ...
The first sequence is 2N where the exponent N is the sum of the geometric series 1 + 2 + ⋯ + k. That sum is well known. You can finish the details. The second sequence is just the sequence of factorials n!. Yes, The first ratio is the base. The second ratio leads to the sum of a geometric series for the exponent.
10. The arithmetic and geometric adjectives come from the Pythagoreans before the Christian Era. Apparently, the expression “geometric progression” comes from the “ geometric mean ” (Euclidean notion) of segments of length a a and b b: it is the length of the side c c of a square whose area is equal to the area of the rectangle of sides ...
4,283 5 38 59. Yes, you can have a geometric sequence with ratio of 1 1. In that case you have another sum formula, ie. S(n) = n ⋅ 2 S (n) = n ⋅ 2 (assuming you are numering from 1 1) – luka5z. Commented Oct 17, 2015 at 17:50.
can a geometric sequence have ratio if 1. 0. Is there a specific formula for geometric sequences? 1.
I have a geometric series with the first term 8 and a common ratio of -3. The last term of this sequence is 52488. I need to find the sum till the nth term. While calculating the nth term for 5248...