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In mathematics, the ratio test is a test (or "criterion") for the convergence of a series
∑ n = 1 ∞ a n , {\displaystyle \sum _{n=1}^{\infty }a_{n},}where each term is a real or complex number and an is nonzero when n is large. The test was first published by Jean le Rond d'Alembert and is sometimes known as d'Alembert's ratio test or as the Cauchy ratio test.
The usual form of the test makes use of the limit
L = lim n → ∞ | a n + 1 a n | . {\displaystyle L=\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|.} |
|
(1) |
The ratio test states that:
It is possible to make the ratio test applicable to certain cases where the limit L fails to exist, if limit superior and limit inferior are used. The test criteria can also be refined so that the test is sometimes conclusive even when L = 1. More specifically, let
R = lim sup | a n + 1 a n | {\displaystyle R=\lim \sup \left|{\frac {a_{n+1}}{a_{n}}}\right|} r = lim inf | a n + 1 a n | {\displaystyle r=\lim \inf \left|{\frac {a_{n+1}}{a_{n}}}\right|} .Then the ratio test states that:
If the limit L in (1) exists, we must have L = R = r. So the original ratio test is a weaker version of the refined one.
Consider the series
∑ n = 1 ∞ n e n {\displaystyle \sum _{n=1}^{\infty }{\frac {n}{e^{n}}}}Applying the ratio test, one computes the limit
L = lim n → ∞ | a n + 1 a n | = lim n → ∞ | n + 1 e n + 1 n e n | = 1 e < 1. {\displaystyle L=\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=\lim _{n\to \infty }\left|{\frac {\frac {n+1}{e^{n+1}}}{\frac {n}{e^{n}}}}\right|={\frac {1}{e}}<1.}Since this limit is less than 1, the series converges.
Consider the series
∑ n = 1 ∞ e n n . {\displaystyle \sum _{n=1}^{\infty }{\frac {e^{n}}{n}}.}Putting this into the ratio test:
L = lim n → ∞ | a n + 1 a n | = lim n → ∞ | e n + 1 n + 1 e n n | = e > 1. {\displaystyle L=\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=\lim _{n\to \infty }\left|{\frac {\frac {e^{n+1}}{n+1}}{\frac {e^{n}}{n}}}\right|=e>1.}Thus the series diverges.
Consider the three series
∑ n = 1 ∞ 1 , {\displaystyle \sum _{n=1}^{\infty }1,} ∑ n = 1 ∞ 1 n 2 , {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}},} ∑ n = 1 ∞ ( − 1 ) n + 1 n . {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}.}The first series (1 + 1 + 1 + 1 + ⋯) diverges, the second one (the one central to the Basel problem) converges absolutely and the third one (the alternating harmonic series) converges conditionally. However, the term-by-term magnitude ratios | a n + 1 a n | {\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|} of the three series are respectively 1 , {\displaystyle 1,} n 2 ( n + 1 ) 2 {\displaystyle {\frac {n^{2}}{(n+1)^{2}}}} and n n + 1 {\displaystyle {\frac {n}{n+1}}} . So, in all three cases, one has that the limit lim n → ∞ | a n + 1 a n | {\displaystyle \lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|} is equal to 1. This illustrates that when L = 1, the series may converge or diverge, and hence the original ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence.
Below is a proof of the validity of the original ratio test.
Suppose that L = lim n → ∞ | a n + 1 a n | < 1 {\displaystyle L=\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|<1} . We can then show that the series converges absolutely by showing that its terms will eventually become less than those of a certain convergent geometric series. To do this, consider a real number r such that L < r < 1 {\displaystyle L<r<1} . This implies that | a n + 1 | < r | a n | {\displaystyle |a_{n+1}|<r|a_{n}|} for sufficiently large n; say, for all n greater than N. Hence | a n + i | < r i | a n | {\displaystyle |a_{n+i}|<r^{i}|a_{n}|} for each n > N and i > 0, and so
∑ i = N + 1 ∞ | a i | = ∑ i = 1 ∞ | a N + i | < ∑ i = 1 ∞ r i | a N | = | a N | ∑ i = 1 ∞ r i = | a N | r 1 − r < ∞ . {\displaystyle \sum _{i=N+1}^{\infty }|a_{i}|=\sum _{i=1}^{\infty }\left|a_{N+i}\right|<\sum _{i=1}^{\infty }r^{i}|a_{N}|=|a_{N}|\sum _{i=1}^{\infty }r^{i}=|a_{N}|{\frac {r}{1-r}}<\infty .}That is, the series converges absolutely.
On the other hand, if L > 1, then | a n + 1 | > | a n | {\displaystyle |a_{n+1}|>|a_{n}|} for sufficiently large n, so that the limit of the summands is non-zero. Hence the series diverges.
As seen in the previous example, the ratio test may be inconclusive when the limit of the ratio is 1. Extensions to the ratio test, however, sometimes allows one to deal with this case.
In all the tests below one assumes that Σan is a sum with positive an. These tests also may be applied to any series with a finite number of negative terms. Any such series may be written as:
∑ n = 1 ∞ a n = ∑ n = 1 N a n + ∑ n = N + 1 ∞ a n {\displaystyle \sum _{n=1}^{\infty }a_{n}=\sum _{n=1}^{N}a_{n}+\sum _{n=N+1}^{\infty }a_{n}}where aN is the highest-indexed negative term. The first expression on the right is a partial sum which will be finite, and so the convergence of the entire series will be determined by the convergence properties of the second expression on the right, which may be re-indexed to form a series of all positive terms beginning at n=1.
Each test defines a test parameter (ρn) which specifies the behavior of that parameter needed to establish convergence or divergence. For each test, a weaker form of the test exists which will instead place restrictions upon limn->∞ρn.
All of the tests have regions in which they fail to describe the convergence properties of Σan. In fact, no convergence test can fully describe the convergence properties of the series. This is because if Σan is convergent, a second convergent series Σbn can be found which converges more slowly: i.e., it has the property that limn->∞ (bn/an) = ∞. Furthermore, if Σan is divergent, a second divergent series Σbn can be found which diverges more slowly: i.e., it has the property that limn->∞ (bn/an) = 0. Convergence tests essentially use the comparison test on some particular family of an, and fail for sequences which converge or diverge more slowly.
Augustus De Morgan proposed a hierarchy of ratio-type tests
The ratio test parameters ( ρ n {\displaystyle \rho _{n}} ) below all generally involve terms of the form D n a n / a n + 1 − D n + 1 {\displaystyle D_{n}a_{n}/a_{n+1}-D_{n+1}} . This term may be multiplied by a n + 1 / a n {\displaystyle a_{n+1}/a_{n}} to yield D n − D n + 1 a n + 1 / a n {\displaystyle D_{n}-D_{n+1}a_{n+1}/a_{n}} . This term can replace the former term in the definition of the test parameters and the conclusions drawn will remain the same. Accordingly, there will be no distinction drawn between references which use one or the other form of the test parameter.
1. d'Alembert's ratio testThe first test in the De Morgan hierarchy is the ratio test as described above.
2. Raabe's testThis extension is due to Joseph Ludwig Raabe. Define:
ρ n ≡ n ( a n a n + 1 − 1 ) {\displaystyle \rho _{n}\equiv n\left({\frac {a_{n}}{a_{n+1}}}-1\right)}(and some extra terms, see Ali, Blackburn, Feld, Duris (none), Duris2)
For the limit version, the series will:
When the above limit does not exist, it may be possible to use limits superior and inferior. The series will:
Defining ρ n ≡ n ( a n a n + 1 − 1 ) {\displaystyle \rho _{n}\equiv n\left({\frac {a_{n}}{a_{n+1}}}-1\right)} , we need not assume the limit exists; if lim sup ρ n < 1 {\displaystyle \limsup \rho _{n}<1} , then ∑ a n {\displaystyle \sum a_{n}} diverges, while if lim inf ρ n > 1 {\displaystyle \liminf \rho _{n}>1} the sum converges.
The proof proceeds essentially by comparison with ∑ 1 / n R {\displaystyle \sum 1/n^{R}} . Suppose first that lim sup ρ n < 1 {\displaystyle \limsup \rho _{n}<1} . Of course if lim sup ρ n < 0 {\displaystyle \limsup \rho _{n}<0} then a n + 1 ≥ a n {\displaystyle a_{n+1}\geq a_{n}} for large n {\displaystyle n} , so the sum diverges; assume then that 0 ≤ lim sup ρ n < 1 {\displaystyle 0\leq \limsup \rho _{n}<1} . There exists R < 1 {\displaystyle R<1} such that ρ n ≤ R {\displaystyle \rho _{n}\leq R} for all n ≥ N {\displaystyle n\geq N} , which is to say that a n / a n + 1 ≤ ( 1 + R n ) ≤ e R / n {\displaystyle a_{n}/a_{n+1}\leq \left(1+{\frac {R}{n}}\right)\leq e^{R/n}} . Thus a n + 1 ≥ a n e − R / n {\displaystyle a_{n+1}\geq a_{n}e^{-R/n}} , which implies that a n + 1 ≥ a N e − R ( 1 / N + ⋯ + 1 / n ) ≥ c a N e − R log ( n ) = c a N / n R {\displaystyle a_{n+1}\geq a_{N}e^{-R(1/N+\dots +1/n)}\geq ca_{N}e^{-R\log(n)}=ca_{N}/n^{R}} for n ≥ N {\displaystyle n\geq N} ; since R < 1 {\displaystyle R<1} this shows that ∑ a n {\displaystyle \sum a_{n}} diverges.
The proof of the other half is entirely analogous, with most of the inequalities simply reversed. We need a preliminary inequality to use in place of the simple 1 + t < e t {\displaystyle 1+t<e^{t}} that was used above: Fix R {\displaystyle R} and N {\displaystyle N} . Note that log ( 1 + R n ) = R n + O ( 1 n 2 ) {\displaystyle \log \left(1+{\frac {R}{n}}\right)={\frac {R}{n}}+O\left({\frac {1}{n^{2}}}\right)} . So log ( ( 1 + R N ) … ( 1 + R n ) ) = R ( 1 N + ⋯ + 1 n ) + O ( 1 ) = R log ( n ) + O ( 1 ) {\displaystyle \log \left(\left(1+{\frac {R}{N}}\right)\dots \left(1+{\frac {R}{n}}\right)\right)=R\left({\frac {1}{N}}+\dots +{\frac {1}{n}}\right)+O(1)=R\log(n)+O(1)} ; hence ( 1 + R N ) … ( 1 + R n ) ≥ c n R {\displaystyle \left(1+{\frac {R}{N}}\right)\dots \left(1+{\frac {R}{n}}\right)\geq cn^{R}} .
Suppose now that lim inf ρ n > 1 {\displaystyle \liminf \rho _{n}>1} . Arguing as in the first paragraph, using the inequality established in the previous paragraph, we see that there exists R > 1 {\displaystyle R>1} such that a n + 1 ≤ c a N n − R {\displaystyle a_{n+1}\leq ca_{N}n^{-R}} for n ≥ N {\displaystyle n\geq N} ; since R > 1 {\displaystyle R>1} this shows that ∑ a n {\displaystyle \sum a_{n}} converges.
3. Bertrand's testThis extension is due to Joseph Bertrand and Augustus De Morgan.
Defining:
ρ n ≡ n ln n ( a n a n + 1 − 1 ) − ln n {\displaystyle \rho _{n}\equiv n\ln n\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\ln n}Bertrand's test asserts that the series will:
For the limit version, the series will:
When the above limit does not exist, it may be possible to use limits superior and inferior. The series will:
This extension probably appeared at the first time by Margaret Martin in 1941. A short proof based on Kummer's test and without technical assumptions (such as existence of the limits, for example) was provided by Vyacheslav Abramov in 2019.
Let K ≥ 1 {\displaystyle K\geq 1} be an integer, and let ln ( K ) ( x ) {\displaystyle \ln _{(K)}(x)} denote the K {\displaystyle K} th iterate of natural logarithm, i.e. ln ( 1 ) ( x ) = ln ( x ) {\displaystyle \ln _{(1)}(x)=\ln(x)} and for any 2 ≤ k ≤ K {\displaystyle 2\leq k\leq K} , ln ( k ) ( x ) = ln ( k − 1 ) ( ln ( x ) ) {\displaystyle \ln _{(k)}(x)=\ln _{(k-1)}(\ln(x))} .
Suppose that the ratio a n / a n + 1 {\displaystyle a_{n}/a_{n+1}} , when n {\displaystyle n} is large, can be presented in the form
a n a n + 1 = 1 + 1 n + 1 n ∑ i = 1 K − 1 1 ∏ k = 1 i ln ( k ) ( n ) + ρ n n ∏ k = 1 K ln ( k ) ( n ) , K ≥ 1. {\displaystyle {\frac {a_{n}}{a_{n+1}}}=1+{\frac {1}{n}}+{\frac {1}{n}}\sum _{i=1}^{K-1}{\frac {1}{\prod _{k=1}^{i}\ln _{(k)}(n)}}+{\frac {\rho _{n}}{n\prod _{k=1}^{K}\ln _{(k)}(n)}},\quad K\geq 1.}(The empty sum is assumed to be 0. With K = 1 {\displaystyle K=1} , the test reduces to Bertrand's test.)
The value ρ n {\displaystyle \rho _{n}} can be presented explicitly in the form
ρ n = n ∏ k = 1 K ln ( k ) ( n ) ( a n a n + 1 − 1 ) − ∑ j = 1 K ∏ k = 1 j ln ( K − k + 1 ) ( n ) . {\displaystyle \rho _{n}=n\prod _{k=1}^{K}\ln _{(k)}(n)\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\sum _{j=1}^{K}\prod _{k=1}^{j}\ln _{(K-k+1)}(n).}Extended Bertrand's test asserts that the series
For the limit version, the series
When the above limit does not exist, it may be possible to use limits superior and inferior. The series
For applications of Extended Bertrand's test see birth–death process.
5. Gauss's testThis extension is due to Carl Friedrich Gauss.
Assuming an > 0 and r > 1, if a bounded sequence Cn can be found such that for all n:
a n a n + 1 = 1 + ρ n + C n n r {\displaystyle {\frac {a_{n}}{a_{n+1}}}=1+{\frac {\rho }{n}}+{\frac {C_{n}}{n^{r}}}}then the series will:
This extension is due to Ernst Kummer.
Let ζn be an auxiliary sequence of positive constants. Define
ρ n ≡ ( ζ n a n a n + 1 − ζ n + 1 ) {\displaystyle \rho _{n}\equiv \left(\zeta _{n}{\frac {a_{n}}{a_{n+1}}}-\zeta _{n+1}\right)}Kummer's test states that the series will:
For the limit version, the series will:
When the above limit does not exist, it may be possible to use limits superior and inferior. The series will
All of the tests in De Morgan's hierarchy except Gauss's test can easily be seen as special cases of Kummer's test:
where the empty product is assumed to be 1. Then,
ρ Kummer = n ∏ k = 1 K ln ( k ) ( n ) a n a n + 1 − ( n + 1 ) + o ( 1 ) = n ∏ k = 1 K ln ( k ) ( n ) ( a n a n + 1 − 1 ) − ∑ j = 1 K ∏ k = 1 j ln ( K − k + 1 ) ( n ) − 1 + o ( 1 ) . {\displaystyle \rho _{\text{Kummer}}=n\prod _{k=1}^{K}\ln _{(k)}(n){\frac {a_{n}}{a_{n+1}}}-(n+1)\left+o(1)=n\prod _{k=1}^{K}\ln _{(k)}(n)\left({\frac {a_{n}}{a_{n+1}}}-1\right)-\sum _{j=1}^{K}\prod _{k=1}^{j}\ln _{(K-k+1)}(n)-1+o(1).}Hence,
ρ Kummer = ρ Extended Bertrand − 1. {\displaystyle \rho _{\text{Kummer}}=\rho _{\text{Extended Bertrand}}-1.}Note that for these four tests, the higher they are in the De Morgan hierarchy, the more slowly the 1 / ζ n {\displaystyle 1/\zeta _{n}} series diverges.
Proof of Kummer's testIf ρ n > 0 {\displaystyle \rho _{n}>0} then fix a positive number 0 < δ < ρ n {\displaystyle 0<\delta <\rho _{n}} . There exists a natural number N {\displaystyle N} such that for every n > N , {\displaystyle n>N,}
δ ≤ ζ n a n a n + 1 − ζ n + 1 . {\displaystyle \delta \leq \zeta _{n}{\frac {a_{n}}{a_{n+1}}}-\zeta _{n+1}.}Since a n + 1 > 0 {\displaystyle a_{n+1}>0} , for every n > N , {\displaystyle n>N,}
0 ≤ δ a n + 1 ≤ ζ n a n − ζ n + 1 a n + 1 . {\displaystyle 0\leq \delta a_{n+1}\leq \zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}.}In particular ζ n + 1 a n + 1 ≤ ζ n a n {\displaystyle \zeta _{n+1}a_{n+1}\leq \zeta _{n}a_{n}} for all n ≥ N {\displaystyle n\geq N} which means that starting from the index N {\displaystyle N} the sequence ζ n a n > 0 {\displaystyle \zeta _{n}a_{n}>0} is monotonically decreasing and positive which in particular implies that it is bounded below by 0. Therefore, the limit
lim n → ∞ ζ n a n = L {\displaystyle \lim _{n\to \infty }\zeta _{n}a_{n}=L} exists.This implies that the positive telescoping series
∑ n = 1 ∞ ( ζ n a n − ζ n + 1 a n + 1 ) {\displaystyle \sum _{n=1}^{\infty }\left(\zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}\right)} is convergent,and since for all n > N , {\displaystyle n>N,}
δ a n + 1 ≤ ζ n a n − ζ n + 1 a n + 1 {\displaystyle \delta a_{n+1}\leq \zeta _{n}a_{n}-\zeta _{n+1}a_{n+1}}by the direct comparison test for positive series, the series ∑ n = 1 ∞ δ a n + 1 {\displaystyle \sum _{n=1}^{\infty }\delta a_{n+1}} is convergent.
On the other hand, if ρ < 0 {\displaystyle \rho <0} , then there is an N such that ζ n a n {\displaystyle \zeta _{n}a_{n}} is increasing for n > N {\displaystyle n>N} . In particular, there exists an ϵ > 0 {\displaystyle \epsilon >0} for which ζ n a n > ϵ {\displaystyle \zeta _{n}a_{n}>\epsilon } for all n > N {\displaystyle n>N} , and so ∑ n a n = ∑ n a n ζ n ζ n {\displaystyle \sum _{n}a_{n}=\sum _{n}{\frac {a_{n}\zeta _{n}}{\zeta _{n}}}} diverges by comparison with ∑ n ϵ ζ n {\displaystyle \sum _{n}{\frac {\epsilon }{\zeta _{n}}}} .
A new version of Kummer's test was established by Tong. See also for further discussions and new proofs. The provided modification of Kummer's theorem characterizes all positive series, and the convergence or divergence can be formulated in the form of two necessary and sufficient conditions, one for convergence and another for divergence.
The first of these statements can be simplified as follows:
The second statement can be simplified similarly:
However, it becomes useless, since the condition ∑ n = 1 ∞ 1 ζ n = ∞ {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\zeta _{n}}}=\infty } in this case reduces to the original claim ∑ n = 1 ∞ a n = ∞ . {\displaystyle \sum _{n=1}^{\infty }a_{n}=\infty .}
Another ratio test that can be set in the framework of Kummer's theorem was presented by Orrin Frink 1948.
Suppose a n {\displaystyle a_{n}} is a sequence in C ∖ { 0 } {\displaystyle \mathbb {C} \setminus \{0\}} ,
This result reduces to a comparison of ∑ n | a n | {\displaystyle \sum _{n}|a_{n}|} with a power series ∑ n n − p {\displaystyle \sum _{n}n^{-p}} , and can be seen to be related to Raabe's test.
A more refined ratio test is the second ratio test: For a n > 0 {\displaystyle a_{n}>0} define:
L 0 ≡ lim n → ∞ a 2 n a n {\displaystyle L_{0}\equiv \lim _{n\rightarrow \infty }{\frac {a_{2n}}{a_{n}}}} |
L 1 ≡ lim n → ∞ a 2 n + 1 a n {\displaystyle L_{1}\equiv \lim _{n\rightarrow \infty }{\frac {a_{2n+1}}{a_{n}}}} |
L ≡ max ( L 0 , L 1 ) {\displaystyle L\equiv \max(L_{0},L_{1})} |
By the second ratio test, the series will:
If the above limits do not exist, it may be possible to use the limits superior and inferior. Define:
L 0 ≡ lim sup n → ∞ a 2 n a n {\displaystyle L_{0}\equiv \limsup _{n\rightarrow \infty }{\frac {a_{2n}}{a_{n}}}} | L 1 ≡ lim sup n → ∞ a 2 n + 1 a n {\displaystyle L_{1}\equiv \limsup _{n\rightarrow \infty }{\frac {a_{2n+1}}{a_{n}}}} | |
ℓ 0 ≡ lim inf n → ∞ a 2 n a n {\displaystyle \ell _{0}\equiv \liminf _{n\rightarrow \infty }{\frac {a_{2n}}{a_{n}}}} | ℓ 1 ≡ lim inf n → ∞ a 2 n + 1 a n {\displaystyle \ell _{1}\equiv \liminf _{n\rightarrow \infty }{\frac {a_{2n+1}}{a_{n}}}} | |
L ≡ max ( L 0 , L 1 ) {\displaystyle L\equiv \max(L_{0},L_{1})} | ℓ ≡ min ( ℓ 0 , ℓ 1 ) {\displaystyle \ell \equiv \min(\ell _{0},\ell _{1})} |
Then the series will:
This test is a direct extension of the second ratio test. For 0 ≤ k ≤ m − 1 , {\displaystyle 0\leq k\leq m-1,} and positive a n {\displaystyle a_{n}} define:
L k ≡ lim n → ∞ a m n + k a n {\displaystyle L_{k}\equiv \lim _{n\rightarrow \infty }{\frac {a_{mn+k}}{a_{n}}}} |
L ≡ max ( L 0 , L 1 , … , L m − 1 ) {\displaystyle L\equiv \max(L_{0},L_{1},\ldots ,L_{m-1})} |
By the m {\displaystyle m} th ratio test, the series will:
If the above limits do not exist, it may be possible to use the limits superior and inferior. For 0 ≤ k ≤ m − 1 {\displaystyle 0\leq k\leq m-1} define:
L k ≡ lim sup n → ∞ a m n + k a n {\displaystyle L_{k}\equiv \limsup _{n\rightarrow \infty }{\frac {a_{mn+k}}{a_{n}}}} | ||
ℓ k ≡ lim inf n → ∞ a m n + k a n {\displaystyle \ell _{k}\equiv \liminf _{n\rightarrow \infty }{\frac {a_{mn+k}}{a_{n}}}} | ||
L ≡ max ( L 0 , L 1 , … , L m − 1 ) {\displaystyle L\equiv \max(L_{0},L_{1},\ldots ,L_{m-1})} | ℓ ≡ min ( ℓ 0 , ℓ 1 , … , ℓ m − 1 ) {\displaystyle \ell \equiv \min(\ell _{0},\ell _{1},\ldots ,\ell _{m-1})} |
Then the series will:
This test is an extension of the m {\displaystyle m} th ratio test.
Assume that the sequence a n {\displaystyle a_{n}} is a positive decreasing sequence.
Let φ : Z + → Z + {\displaystyle \varphi :\mathbb {Z} ^{+}\to \mathbb {Z} ^{+}} be such that lim n → ∞ n φ ( n ) {\displaystyle \lim _{n\to \infty }{\frac {n}{\varphi (n)}}} exists. Denote α = lim n → ∞ n φ ( n ) {\displaystyle \alpha =\lim _{n\to \infty }{\frac {n}{\varphi (n)}}} , and assume 0 < α < 1 {\displaystyle 0<\alpha <1} .
Assume also that lim n → ∞ a φ ( n ) a n = L . {\displaystyle \lim _{n\to \infty }{\frac {a_{\varphi (n)}}{a_{n}}}=L.}
Then the series will: