How to Determine Convergence of Infinite Series

Perform the divergence test. This test determines whether the series u k {\displaystyle u_{k}} u_{{k}} is divergent or not, where k ∈ Z . {\displaystyle k\in \mathbb {Z} .} k\in {\mathbb {Z}}. If lim k → ∞ u k ≠ 0 , {\displaystyle \lim _{k\to \infty }u_{k}\neq 0,} \lim _{{k\to \infty }}u_{{k}}\neq 0, then u k {\displaystyle u_{k}} u_{{k}} diverges. The inverse is not true. If the limit of a series is 0, that does not necessarily mean that the series converges. We must do further checks.

Look for geometric series. Geometric series are series of the form r k , {\displaystyle r^{k},} r^{{k}}, where r {\displaystyle r} r is the ratio between two adjacent numbers in the series. These series are very easy to recognize and determine the convergence of. If | r | < 1 , {\displaystyle |r|<1,} |r|<1, then r k {\displaystyle r^{k}} r^{{k}} converges. If | r | ≥ 1 , {\displaystyle |r|\geq 1,} |r|\geq 1, then r k {\displaystyle r^{k}} r^{{k}} diverges. If r = − 1 , {\displaystyle r=-1,} r=-1, then the test is inconclusive. Use the alternating series test. For convergent geometric series, you can find the sum of the series as a 1 − r {\displaystyle {\frac {a}{1-r}}} {\displaystyle {\frac {a}{1-r}}} where a {\displaystyle a} a is the first term of the sequence.

Look for p-series. P-series are series of the form 1 k p . {\displaystyle {\frac {1}{k^{p}}}.} {\frac {1}{k^{{p}}}}. They are sometimes called "hyperharmonic" series for the way they generalize the harmonic series, of which p = 1. {\displaystyle p=1.} p=1. If p > 1 , {\displaystyle p>1,} p>1, then the series converges. If 0 < p ≤ 1 , {\displaystyle 0

Perform the integral test. This test works best when f {\displaystyle f} f is easy to integrate. Note that f {\displaystyle f} f must be decreasing, or the series automatically diverges. Given a decreasing, continuous function f {\displaystyle f} f where u k = f ( k ) {\displaystyle u_{k}=f(k)} u_{{k}}=f(k) for all k ≥ a , {\displaystyle k\geq a,} k\geq a, then u k {\displaystyle u_{k}} u_{{k}} and ∫ a ∞ f ( x ) d x {\displaystyle \int _{a}^{\infty }f(x)\mathrm {d} x} \int _{{a}}^{{\infty }}f(x){\mathrm {d}}x both converge or both diverge. In other words, we can construct a continuous function out of a discrete series, where the terms between the series and the function are equal to one another. Then, we can simply evaluate the integral to check for divergence. If it is divergent, then the series is divergent as well. Going back to the harmonic series, this series can be represented by the function f ( x ) = 1 x . {\displaystyle f(x)={\frac {1}{x}}.} f(x)={\frac {1}{x}}. Since ∫ 1 ∞ 1 x d x = ln ⁡ ( ∞ ) − ln ⁡ ( 1 ) = ∞ {\displaystyle \int _{1}^{\infty }{\frac {1}{x}}\mathrm {d} x=\ln(\infty )-\ln(1)=\infty } \int _{{1}}^{{\infty }}{\frac {1}{x}}{\mathrm {d}}x=\ln(\infty )-\ln(1)=\infty (because the logarithmic function is unbounded), the integral test is yet another way of showing the divergence of this series.

Perform the alternating series test for alternating series. These series usually contain a ( − 1 ) k {\displaystyle (-1)^{k}} (-1)^{{k}} term in it. All other tests in this article pertain to series with all positive terms. If a k > 0 {\displaystyle a_{k}>0} a_{{k}}>0 for a sufficiently large k , {\displaystyle k,} k, then a k {\displaystyle a_{k}} a_{{k}} converges if the following two conditions hold. a k ≥ a k + 1 {\displaystyle a_{k}\geq a_{k+1}} a_{{k}}\geq a_{{k+1}} lim k → ∞ a k = 0 {\displaystyle \lim _{k\to \infty }a_{k}=0} \lim _{{k\to \infty }}a_{{k}}=0 Put more simply, if you have an alternating series, ignore the signs and check if each term is less than the previous term. Then check if the limit of the series goes to 0. It is useful to note that series that converge via the alternating series test, but diverge when the ( − 1 ) k {\displaystyle (-1)^{k}} (-1)^{{k}} is removed, are deemed conditionally convergent. The alternating harmonic series ( − 1 ) k − 1 k {\displaystyle {\frac {(-1)^{k-1}}{k}}} {\frac {(-1)^{{k-1}}}{k}} is one such example, whose sum is ln ⁡ 2. {\displaystyle \ln 2.} \ln 2.

Perform the ratio test. This test is useful for expressions with factorials or powers in them. Given an infinite series u k , {\displaystyle u_{k},} u_{{k}}, find u k + 1 {\displaystyle u_{k+1}} u_{{k+1}} and compute | u k + 1 u k | . {\displaystyle \left\vert {\frac {u_{k+1}}{u_{k}}}\right\vert .} \left\vert {\frac {u_{{k+1}}}{u_{{k}}}}\right\vert . Now let ρ = lim k → ∞ | u k + 1 u k | . {\displaystyle \rho =\lim _{k\to \infty }\left\vert {\frac {u_{k+1}}{u_{k}}}\right\vert .} \rho =\lim _{{k\to \infty }}\left\vert {\frac {u_{{k+1}}}{u_{{k}}}}\right\vert . The series converges (even absolutely) if ρ < 1 {\displaystyle \rho <1} \rho <1, diverges if ρ > 1 {\displaystyle \rho >1} \rho >1 or ρ = ∞ , {\displaystyle \rho =\infty ,} \rho =\infty , and is inconclusive if ρ = 1. {\displaystyle \rho =1.} \rho =1. Note that the ratio test doesn't work if u k = 0 {\displaystyle u_{k}=0} u_{k}=0 for any k {\displaystyle k} k. In this case, the series has to be rewritten in a way that no zeros are added, or if that is too much work, the root test has to be used.

Perform the root test. The root test is a variant of the ratio test, where ρ = lim k → ∞ | u k | k . {\displaystyle \rho =\lim _{k\to \infty }{\sqrt[{k}]{\vert u_{k}\vert }}.} \rho =\lim _{{k\to \infty }}{\sqrt[ {k}]{\vert u_{{k}}\vert }}. The same criteria from the ratio test are used for the root test. A stronger version of the root test uses ρ = lim sup k → ∞ | u k | k . {\displaystyle \rho =\limsup _{k\to \infty }{\sqrt[{k}]{\vert u_{k}\vert }}.} \rho =\limsup _{{k\to \infty }}{\sqrt[ {k}]{\vert u_{{k}}\vert }}.. The criteria are the same, but the limit superior might exist while the limit doesn't. This version of the test also works in those cases. The root test is strictly stronger than the ratio test, especially with the limit superior version. There are series for which the ratio test is inconclusive, but the root test is conclusive, even though they work in similar ways. Note that the root of the absolute value of u k {\displaystyle u_{k}} u_{k} is taken.

Perform the limit comparison test. This test involves choosing a sufficient series b k {\displaystyle b_{k}} b_{{k}} for which you know the convergence/divergence of, and compares it to a series a k {\displaystyle a_{k}} a_{{k}} through a limit. This test is often used in evaluating the convergence of series defined by rational expressions. Let ρ = lim k → ∞ a k b k . {\displaystyle \rho =\lim _{k\to \infty }{\frac {a_{k}}{b_{k}}}.} \rho =\lim _{{k\to \infty }}{\frac {a_{{k}}}{b_{{k}}}}. Then the series both converge if ρ {\displaystyle \rho } \rho is finite, or both diverge if ρ = ± ∞ . {\displaystyle \rho =\pm \infty .} \rho =\pm \infty . For example, if you were given a series 1 k 3 + 2 k + 1 , {\displaystyle {\frac {1}{k^{3}+2k+1}},} {\frac {1}{k^{{3}}+2k+1}}, then it makes sense to compare it to 1 k 3 , {\displaystyle {\frac {1}{k^{3}}},} {\frac {1}{k^{{3}}}}, as the highest-ordered term increases/falls off the quickest, and you know the latter is convergent via the p-series test.

Perform the comparison test. This test is generally cumbersome, so use it as a last resort. Given two positive term series a k {\displaystyle a_{k}} a_{{k}} and b k , {\displaystyle b_{k},} b_{{k}}, and the kth term of a {\displaystyle a} a is less than the kth term of b , {\displaystyle b,} b, then the following are true. If the bigger series b k {\displaystyle b_{k}} b_{{k}} converges, then the smaller series a k {\displaystyle a_{k}} a_{{k}} converges as well, since b k > a k . {\displaystyle b_{k}>a_{k}.} b_{{k}}>a_{{k}}. If the smaller series a k {\displaystyle a_{k}} a_{{k}} diverges, then the bigger series b k {\displaystyle b_{k}} b_{{k}} diverges as well, since a k < b k . {\displaystyle a_{k}