Convergenceand divergence series tests

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Accordingto D’Angelo and West (2000), a series is termed to be convergentonly when it approaches some limit. In normal understanding, aninfinite series  isconvergent if the order of partial sums is convergent. In the sameway, a series is said to be divergent in case the partial sumssequence is divergent. For instance, if  and  indicatea series that is convergent, then their respective sums areconvergent i.e  and .The convergence nature of any series is not affected by deleting afinite number of terms from the start of the series. Terms that areconstant in the denominator of the sequence can always be omitteddevoid of impacting the convergence. There are various tests that areused to test the convergence of a series. This paper highlights theuse of integral and root tests of convergence.

Roottest

Assumingwe have a series ∑a_{n}, we define

Forroot test if L is less than 1, then the series is convergent (L<1,convergence), when L is greater than 1 (L>1), the series isdivergent and when L equals to 1 (L=1), the series is inconclusive ieit may be conditionally convergent, divergent or absolutelyconvergent.

Integraltest

Justas the root test, integral test determines whether a series isdivergent, convergent or inconclusive. Assuming we have a series ∑a_{n},

If*f*is a positive, continous and decreasing function on interval (1, ∞)with f(n) = a_{n}thenthe series∑a_{n} diverges if and only ifthe improper integral diverges. The reverse is also true (Zagarrel, 2008).

Althoughthe root test and the integral test are too different from eachother, there are times when either can be used to determine theconvergence of a series and times when only one can and the other isinconclusive. We shall look at examples in which both the tests canbe used and in which only one test can be applied and not the other.

Examplesof series whose convergence/divergence can be determined by usingboth test methods

Example1:we can use both integral and root test to determine if is divergent or convergent

Usingroot test

Wehave

Andit is clear that when n is approaching infinite

Therefore

ThusL is equal to 1 implying that the series is inconclusive.

Usingintegral test

== ispositive continuous and decreasing

=lim(b appro ∞)

Theseries cannot be integrated fully, thus the series is inconclusive

Example2

Usingintegral test

F(x)= x/(x^{2}+4)^{1/2}is positive continuous and decreasing

=lim(b appro ∞√x^{2}+4

Thusthe series diverges

Usingroot test

Wesquare everything to get ∑ k^{2}/K^{2}+4

Thendivide each by k^{2}givingus 1/1 + 4/ k^{2}

Sincek approaches infinite the result for lim(bappro ∞1/1 + 4/ k^{2}>1 hence the series is divergent

Example3

∑ ke^{-2k2}

Forthis series, an integral test shows that it converges

Thatis f(xe^{-2×2}is continuous and positive for x=1

Integratingit, we get e^{-2×2}+ xe^{-2×2}(-4x)= e-2×2 (1-4x^{2})which is less than zero making the function decreasing

Integratingfrom 1 to infinite the values of x, we get e^{-2}/4which is a clear indication that the series converges

Forroot test:

Wefind the square root of the series and get ∑x^{1/2}e^{x2}

ThereforeL= lim(x appro ∞)∑x^{1/2}e^{x2} which is less than 1 thus the series is convergent

Examplesof series that whose convergence/divergence can be determined byusing one test method chosen, but another one inconclusive

Example1

for this example, if we use integral test, we understand that

=e^{-x}is positive continuous and decreasing

=lim(b appro ∞)= lim(b appro ∞)(e^{-2}– e^{-b})

=e^{-2}which clearly indicates that the series is convergent

Usingroot test

L= lim(k appro ∞)1/e^{k}

We1/1 = 1 therefore, using root test, the series is inconclusive

Example2

Usingroot test L = lim(k appro ∞)k√()

=lim(k appro ∞)(√k)^{2}/2

=½ < 1 hence the series is convergent

Usingintegral test

Wehave

=ispositive continuous but not decreasing hence the series isinconclusive using the integral test.

Reference

Zegarelli,M. (2008). *CalculusII for dummies*.Hoboken, NJ: Wiley Pub.