Partial Sums of Infinite Series
Key Questions
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Answer:
#195/16# Explanation:
When dealing with a sum, you have a sequence that generates the terms. In this case, you have the sequence
#a_n = (3/2)^n# Which means that
#n# -th term is generates by raising#3/2# to the#n# -th power.Moreover, the
#n# -th partial sum means to sum the first#n# terms from the sequence.So, in your case, you're looking for
#a_1+a_2+a_3+a_4# , which means#3/2 + (3/2)^2 + (3/2)^3 + (3/2)^4# You may compute each term, but there is a useful formula:
#sum_{i=1}^n k^i= \frac{k^{n+1}-1}{k-1}# So, in your case
#sum_{i=0}^4 (3/2)^i= \frac{(3/2)^{5}-1}{3/2-1} = 211/16# Except you are not including
#a_0 = (3/2)^0 = 1# in your sum, so we must subtract it:#sum_{i=0}^4 (3/2)^i = sum_{i=1}^4 (3/2)^i - 1 = 211/16 - 1 = 195/16# -
A partial sum
#S_n# of an infinite series#sum_{i=1}^{infty}a_i# is the sum of the first n terms, that is,
#S_n=a_1+a_2+a_3+cdots+a_n=sum_{i=1}^na_i# .
Questions
Tests of Convergence / Divergence
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Geometric Series
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Nth Term Test for Divergence of an Infinite Series
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Direct Comparison Test for Convergence of an Infinite Series
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Ratio Test for Convergence of an Infinite Series
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Integral Test for Convergence of an Infinite Series
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Limit Comparison Test for Convergence of an Infinite Series
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Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series
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Infinite Sequences
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Root Test for for Convergence of an Infinite Series
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Infinite Series
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Strategies to Test an Infinite Series for Convergence
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Harmonic Series
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Indeterminate Forms and de L'hospital's Rule
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Partial Sums of Infinite Series