are telescoping series always convergent?
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In coursework, it's conventional to conclude that oscillating series like this one diverge, whereas it's conventional to conclude that other telescoping series like the ones Sal shows converge (but if you group them so that each number is paired with one of a different size and either always greater or always less, you ...
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Answer Expert Verified Yes its. Its only one posibility to not have linear function. If we have for example x=2, the graph will be vertical line.
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Question: (a) Can The Graph Of Y = F(x) Intersect A Vertical Asymptote? Yes No Can It Intersect A Horizontal Asymptote?
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Answer and Explanation: No, complementary angles are not always congruent.
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The nth partial sum of this series is sn = ( 1 - 1 2 ) + (1 2 - 1 3 ) + (1 3 - 1 4 ) + (1 4 - 1 5 ) + ··· + (1 n - 1 n + 1 ) = 1 - 1 n + 1 . sn = 1, the series converges with sum 1. arn = a + ar + ar2 + ar3 + ··· , where a is a constant and r is called the ratio of the series.
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O(1) and Θ(1) aren't necessarily the same if you are talking about functions over real numbers. ... This function is O(1) because for any n ≥ 1, f(n) ≤ 1.
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['Average is the easiest method to calculate.', 'It should be rigidly defined. ... ', 'It should be representative of the entire series. ... ', 'It should have capable of further algebraic treatment. ... ', 'It should not be affected by the extreme values of the observation or series.']
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1 Answer. The radius of convergence is usually required to find the interval of convergence. While the radius gives us the number of values where the series converges, the interval gives us the exact values of where the series converges and doesn't.
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In general, if function is increasing then left endpoint approximation underestimates value of integral, while right endpoint approximation overestimates it. If function is decreasing then left endpoint approximation overestimates value of integral, while right endpoint approximation underestimates it.
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Answer and Explanation: No, the difference of two positive rational numbers is not always positive.
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Since the exponential function and the sine are defined for all real x, y is defined for all real x, so there are no vertical asymptotes. ... However, since the function crosses the x - axis 4 times between each integer, this is an example of a function which will cross its asymptote an infinite number of times.
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Note: it is true that every bounded sequence contains a convergent subsequence, and furthermore, every monotonic sequence converges if and only if it is bounded. Added See the entry on the Monotone Convergence Theorem for more information on the guaranteed convergence of bounded monotone sequences.
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yes. It is tree based and thus sensitive to order of values but not actual values. Outliers in target variable are another matter. With many loss functions (such as RMSE/L2) you are necessarily sensitive to outliers.
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In a complete metric space, every Cauchy sequence is convergent. This is because it is the definition of Complete metric space. ... This shows that every Cauchy sequence in converges to a point in , so is a complete metric space.
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Squeezing Theorem. The sequence {an} is bounded if there is a number M>0 such that |an|≤M for every positive n. Every convergent sequence is bounded. Every unbounded sequence is divergent.
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Theorem 2.4: Every convergent sequence is a bounded sequence, that is the set {xn : n ∈ N} is bounded. Remark : The condition given in the previous result is necessary but not sufficient. For example, the sequence ((−1)n) is a bounded sequence but it does not converge.
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['It should be easy to calculate & simple to understand.', 'It should be based on all the observations of the series.', 'It should be rigidly defined.', 'It should not be affected by extreme values.', 'It should not be unduly affected by sampling fluctuations.']