Infinite Series
Key Questions
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Answer:
See below
Explanation:
There are different types of series, to what use different methods of evaluating
For example a converging geometric series:
a+ ar + ar^2 + ar^3 + ... + ar^k = sum_(n=1) ^(k) ar^(n-1) where
sum_(n=1) ^(k) ar^(n-1) = (a(1-r^k)) / (1-r) Assuming
|r| < 1 we can letk to oo for infinite series to be evaluated ...lim_(k to oo ) sum_(n=1) ^(k) ar^(n-1) = lim_(k to oo ) ( a(1-r^k) )/(1-k) as
k to oo ,r^k to 0 as|r|< 1 => sum_(n=1) ^(oo) ar^(n-1) = a/(1-r) but there are other series what can be approached with tricks!
Take
1/6 + 1/12 + 1/20 + 1/30 + ... After consideration we can recognise this is the same as...
(1/2 - 1/3 ) + ( 1/3 - 1/4) + (1/4 - 1/5 ) + ... (1/2 cancel(- 1/3) ) + (cancel( 1/3) cancel(- 1/4)) + (cancel(1/4) cancel(- 1/5) ) + ... =1/2 There are also other infinite series that you can remember, and may be able to prove, a like:
e^x = 1 + x + x^2 /(2!) + x^3 / (3!) + ... = sum_(n=0) ^oo x^n / (n!) There are many others, where there insist one set way of computing infinite series, there are many!
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I believe that it is the same as an alternating series. If that is the case, then an oscillating series is a series of the form:
sum_{n=0}^infty (-1)^n b_n , whereb_n ge 0 .For example, the alternating harmonic series
sum_{n=1}^infty{(-1)^n}/n is a convergent alternating series.
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Here is an example of a telescoping series
sum_{n=1}^infty(1/n-1/{n+1})
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+cdots
As you can see above, terms are shifted with some overlapping terms, which reminds us of a telescope. In order to find the sum, we will its partial sumS_n first.
S_n=(1/1-1/2)+(1/2-1/3)+cdots+(1/n-1/{n+1})
by cancelling the overlapping terms,
=1-1/{n+1}
Hence, the sume of the infinite series can be found by
sum_{n=1}^infty(1/n-1/{n+1})=lim_{n to infty}S_n=lim_{n to infty}(1-1/{n+1})=1 -
It is very tough to answer such a general question, but I will give it a shot. Infinite series allow us to add up infinitely many terms, so it is suitable for representing something that keeps on going forever; for example, a geometric series can be used to find a fraction equivalent to any given repeating decimal such as:
3.333... by splitting into individual decimals,
=3+0.3+0.03+0.003+cdots by rewriting into a form of geometric series,
=3+3(1/10)+3(1/10)^2+3(1/10)^3+cdots by using the formula for the sum of geometric series,
=3/{1-1/10}=10/3 The knowledge of geometric series helped us find the fraction fairly easily. I hope that this was helpful.
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